\(\left\{{}\begin{matrix}5x-6y=17\\9x-y=7\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}5x-6\left(9x-7\right)=17\\y=9x-7\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}5x-54x+42=17\\y=9x-7\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}-49x=-25\\y=9x-7\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=\frac{25}{49}\\y=9\left(\frac{25}{49}\right)-7=\frac{-118}{49}\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x,y\right)=\left(\frac{25}{49};\frac{-118}{49}\right)\) .

a) \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\) (1)

\(\Leftrightarrow9x-7=\sqrt{\left(7x+5\right)\left(7x+5\right)}\)

\(\Leftrightarrow9x-\sqrt{\left(7x+5\right)\left(7x+5\right)}=7\)

\(\Leftrightarrow9x-\sqrt{\left(7x+5\right)^2}=7\)

\(\Leftrightarrow9x-\left|7x+5\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}9x-\left(7x+5\right)=7\left(đk:7x+5\ge0\right)\\9x-\left[-\left(7x+5\right)\right]=7\left(đk:7x+5< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\left(đk:x\ge-\dfrac{5}{7}\right)\\x=\dfrac{1}{8}\left(đk:x< -\dfrac{5}{7}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x\in\varnothing\end{matrix}\right.\)

\(\Leftrightarrow x=6\)

Vậy tập nghiệm phương trình (1) là \(S=\left\{6\right\}\)

b) \(\sqrt{4x-20}+3\sqrt{\dfrac{x+5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\) (2)

\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3\cdot\dfrac{\sqrt{x+5}}{3}-\dfrac{1}{3}\cdot\sqrt{9\left(x-5\right)}=4\)

\(\Leftrightarrow\sqrt{4}\sqrt{x-5}+\sqrt{x+5}-\dfrac{1}{3}\cdot\sqrt{9}\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x+5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x+5}-\sqrt{x-5}=4\)

\(\Leftrightarrow\sqrt{x-5}+\sqrt{x+5}=4\)

\(\Leftrightarrow\sqrt{x-5}=4-\sqrt{x+5}\)

\(\Leftrightarrow x-5=\left(4-\sqrt{x+5}\right)^2\)

\(\Leftrightarrow x-5=16-8\sqrt{x+5}+x+5\)

\(\Leftrightarrow-5=16-8\sqrt{x+5}+5\)

\(\Leftrightarrow-5=21-8\sqrt{x+5}\)

\(\Leftrightarrow8\sqrt{x+5}=21+5\)

\(\Leftrightarrow8\sqrt{x+5}=26\)

\(\Leftrightarrow\sqrt{x+5}=\dfrac{13}{4}\)

\(\Leftrightarrow x+5=\dfrac{169}{16}\)

\(\Leftrightarrow x=\dfrac{169}{16}-5\)

\(\Leftrightarrow x=\dfrac{89}{16}\)

Vậy tập nghiệm phương trình (2) là \(S=\left\{\dfrac{89}{16}\right\}\)

Nick cũ không đi giải lấy nick mới giải làm gì vậy Tuấn Anh Phan Nguyễn ? :D

a: \(\Leftrightarrow\dfrac{2x-3}{x-1}=4\)

=>4x-4=2x-3

=>2x=1

hay x=1/2

b: \(\Leftrightarrow\sqrt{\dfrac{2x-3}{x-1}}=2\)

=>(2x-3)=4x-4

=>4x-4=2x-3

=>2x=1

hay x=1/2(nhận)

c: \(\Leftrightarrow\sqrt{2x+3}\left(\sqrt{2x-3}-2\right)=0\)

=>2x+3=0 hoặc 2x-3=4

=>x=-3/2 hoặc x=7/2

e: \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

=>căn (x-5)=2

=>x-5=4

hay x=9

a)

ĐKXĐ: \(x> \frac{-5}{7}\)

Ta có: \(\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)

\(\Rightarrow 9x-7=\sqrt{7x+5}.\sqrt{7x+5}=7x+5\)

\(\Rightarrow 2x=12\Rightarrow x=6\) (hoàn toàn thỏa mãn)

Vậy......

b) ĐKXĐ: \(x\geq 5\)

\(\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)

\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}\sqrt{9}.\sqrt{x-5}=4\)

\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

\(\Leftrightarrow 2\sqrt{x-5}=4\Rightarrow \sqrt{x-5}=2\Rightarrow x-5=2^2=4\Rightarrow x=9\)

(hoàn toàn thỏa mãn)

Vậy..........

c) ĐK: \(x\in \mathbb{R}\)

Đặt \(\sqrt{6x^2-12x+7}=a(a\geq 0)\Rightarrow 6x^2-12x+7=a^2\)

\(\Rightarrow 6(x^2-2x)=a^2-7\Rightarrow x^2-2x=\frac{a^2-7}{6}\)

Khi đó:

\(2x-x^2+\sqrt{6x^2-12x+7}=0\)

\(\Leftrightarrow \frac{7-a^2}{6}+a=0\)

\(\Leftrightarrow 7-a^2+6a=0\)

\(\Leftrightarrow -a(a+1)+7(a+1)=0\Leftrightarrow (a+1)(7-a)=0\)

\(\Rightarrow \left[\begin{matrix} a=-1\\ a=7\end{matrix}\right.\) \(\Rightarrow a=7\) vì \(a\geq 0\)

\(\Rightarrow 6x^2-12x+7=a^2=49\)

\(\Rightarrow 6x^2-12x-42=0\Leftrightarrow x^2-2x-7=0\)

\(\Leftrightarrow (x-1)^2=8\Rightarrow x=1\pm 2\sqrt{2}\)

(đều thỏa mãn)

Vậy..........

\(a,\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)\(ĐKXĐ:x\ge-\frac{5}{7}\)

\(\Leftrightarrow9x-7=7x+5\)

\(\Leftrightarrow9x-7x=5+7\)

\(\Leftrightarrow2x=12\)

\(\Leftrightarrow x=6\)

\(b,\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)

\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3.\frac{\sqrt{x-5}}{\sqrt{9}}-\frac{1}{3}\sqrt{9\left(x-5\right)}=4\)

\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

\(\Leftrightarrow\sqrt{x-5}\left(2+1-1\right)=4\)

\(\Leftrightarrow2\sqrt{x-5}=4\)

\(\Leftrightarrow\sqrt{x-5}=2\)

\(\Leftrightarrow x-5=4\)

\(\Leftrightarrow x=9\)

\(PT\Leftrightarrow\left(x+3\right)^2+\left(\sqrt{3x+10}-1\right)^2=0\)

hichic T_T!!! vậy mà cứ chăm chăm đặt ẩn phụ T_T!!!!! Mơn nhak ^^!

Bài theo tôi tương đối đơn giản, tôi sẽ làm ngắn gọn thôi.

Ta viết hệ phương trình trên thành :

\(\left\{{}\begin{matrix}\left(a-2\right)^3=1-3a\left(1\right)\\\left(b-1\right)^3=2-3b\left(2\right)\end{matrix}\right.\)

Lấy (1) trừ (2) ta được :

\(\left(a-2\right)^3-\left(b-1\right)^3=3-3\left(a-b\right)\\ \Rightarrow\left(a-b-1\right)\left[\left(a-2\right)^2+\left(a-2\right)\left(b-1\right)+\left(b-1\right)^2+3\right]=0\)

\(\Rightarrow a-b=1\Rightarrow\left(a-b\right)^{2014}=1\\ Vậy...........\)

Nhầm nhé, của câu trên.

nãy giải rồi

\(\hept{\begin{cases}7x-3y=4\\4x+y=5\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}7x-3y=4\\12x+3y=15\end{cases}}\)

Cộng vế ta được :

\(7x-3y+12x+3y=4+15\)

\(\Leftrightarrow19x=19\)

\(\Leftrightarrow x=1\)

Khi đó : \(7-3y=4\Leftrightarrow y=1\)

Vậy \(x=y=1\)

a) \(x^4-9x^2+20=0\)

\(\Leftrightarrow x^4-4x^2-5x^2+20=0\)

\(\Leftrightarrow x^2\left(x^2-4\right)-5\left(x^2-4\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-4=0\\x^2-5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x\in\left\{\pm2\right\}\\x\in\left\{\pm\sqrt{5}\right\}\end{cases}}\)

Vậy....

\(x^4-9x^2+20=0\)

\(\Leftrightarrow x^4-4x^2-5x^2+20=0\)

\(\Leftrightarrow x^2\left(x^2-4\right)-5\left(x^2-4\right)=0\)

\(\Leftrightarrow\left(x^2-5\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-5=0\\x^2-4=0\end{cases}}\Leftrightarrow x\in\left\{\pm2\right\}\)