9) \(\dfrac{x}{4}=\dfrac{9}{x}\)

Theo định nghĩa về hai phân số bằng nhau, ta có:

\(4\cdot9=x^2\\ 36=x^2\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

8)

\(x:\dfrac{5}{3}+\dfrac{1}{3}=-\dfrac{2}{5}\\ x:\dfrac{5}{3}=-\dfrac{2}{5}+\dfrac{1}{3}\\ x:\dfrac{5}{3}=-\dfrac{1}{15}\\ x=\dfrac{1}{15}\cdot\dfrac{5}{3}\\ x=\dfrac{1}{9}\)

7)

\(2x-16=40+x\\ 2x-x=40+16\\ x\left(2-1\right)=56\\ x=56\)

6)

\(1\dfrac{1}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}-\dfrac{3}{2}=-7-x\\ -7-x=0\\ x=-7-0\\ x=-7\)

5)

\(3\dfrac{1}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{7}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{7}{2}-\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{17}{6}\\ x=\dfrac{17}{6}:\dfrac{1}{2}\\ x=\dfrac{17}{3}\)

4)

\(x\cdot\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

3)

\(\left(\dfrac{2x}{5}+2\right):\left(-4\right)=-1\dfrac{1}{2}\\ \left(\dfrac{2x}{5}+2\right):\left(-4\right)=-\dfrac{3}{2}\\ \dfrac{2x}{5}+2=-\dfrac{3}{2}\cdot\left(-4\right)\\ \dfrac{2x}{5}+2=6\\ \dfrac{2x}{5}=6-2\\ \dfrac{2x}{5}=4\\ 2x=4\cdot5\\ 2x=20\\ x=20:2\\ x=10\)

2)

\(\dfrac{1}{3}+\dfrac{1}{2}:x=-0,25\\ \dfrac{1}{3}+\dfrac{1}{2}:x=-\dfrac{1}{4}\\ \dfrac{1}{2}:x=-\dfrac{1}{4}-\dfrac{1}{3}\\ \dfrac{1}{2}:x=-\dfrac{7}{12}\\ x=\dfrac{1}{2}:-\dfrac{7}{12}\\ x=-\dfrac{6}{7}\)

1)

\(\dfrac{4}{3}+x=\dfrac{2}{15}\\ x=\dfrac{2}{15}-\dfrac{4}{3}x=-\dfrac{6}{5}\)

\(8-12x+6x^2-x^3\)

\(=\left(2-x\right)^3\)

\(125x^3-75x^2+15x-1\)

\(=\left(5x-1\right)^3\)

\(x^2-xz-9y^2+3yz\)

\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)

\(=\left(x-3y\right)\left(x+3y-z\right)\)

\(x^3-x^2-5x+125\)

\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)

\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)

\(=\left(x+5\right)\left(x^2-6x+25\right)\)

\(x^3+2x^2-6x-27\)

\(=x^3+5x^2+9x-3x^2-15x-27\)

\(=x\left(x^2+5x+9\right)-3\left(x^2+5x+9\right)\)

\(=\left(x-3\right)\left(x^2+5x+9\right)\)

\(12x^3+4x^2-27x-9\)

\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)

\(=\left(3x+1\right)\left(4x^2-9\right)\)

\(=\left(3x+1\right)\left(2x-3\right)\left(2x+3\right)\)

\(4x^4+4x^3-x^2-x\)

\(=4x^3\left(x+1\right)-x\left(x+1\right)\)

\(=x\left(x+1\right)\left(4x^2-1\right)\)

\(=x\left(x+1\right)\left(2x-1\right)\left(2x+1\right)\)

hhhhhhhhhhhhhhhhhhh

a,\(x^4:9=9\\ \Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

b,\(x^2-30=6\\ \Leftrightarrow x^2=36\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)

c,\(x^2:10=10\Leftrightarrow x^2=\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

x⁴ : 9 = 9

x⁴ = 9 . 9

x⁴ = 81

x⁴ = 3⁴

=> x = 3

x² - 30 = 6

x² = 6 + 30

x² = 36

x² = 6²

=> x = 6

x² : 10 = 10

x² = 10 . 10

x² = 100

x² = 10²

=> x = 10

\(\left(\frac{2}{9}x-\frac{4}{9}\right).\left(\frac{1}{3}+\frac{-4}{7}:x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{9}x-\frac{4}{9}=0\\\frac{1}{3}+\frac{-4}{7}:x=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{9}.x=\frac{4}{9}\\\frac{4}{7x}=\frac{1}{3}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{12}{7}\end{cases}}\)

1. ta có: \(\sqrt{\dfrac{4}{9}-\sqrt{\dfrac{25}{36}}}=\sqrt{\dfrac{4}{9}-\dfrac{5}{6}}=\sqrt{-\dfrac{7}{18}}\)

Mà \(-\dfrac{7}{18}\) là số âm \(\Rightarrow\) Bài toán không có kết quả.

2. Ta có:

\(\left(x-1\right)^2=\dfrac{9}{16}\)

\(\Rightarrow\left(x-1\right)^2=\left(\dfrac{3}{4}\right)^2\)

\(\Rightarrow x-1=\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{3}{4}+1\)

\(\Rightarrow x=1\dfrac{3}{4}\)

Vậy \(x=1\dfrac{3}{4}\)

Câu 2 không phải toán lớp 6 mà bạn.

Ta có: \(x=\sqrt{x}\)

\(\Rightarrow x=1\)

Vậy \(x=1\)

Bạn Trần Đăng Nhất làm thiếu nha:

\(x=\sqrt{x}=>x^2=\left(\sqrt{x}\right)^2\)

\(=>x^2=x=>x^2-x=0\)

\(=>x\left(x-1\right)=0\)

\(=>\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Vậy có 2 giá trị của x là 0 và 1..

CHÚC BẠN HỌC TỐT.....

tìm x a)
\(\dfrac{7}{2}\)-\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{4}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{4}\) + \(\dfrac{7}{2}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{12}+\dfrac{7}{12}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-12}{12}=1\)
\(x+\dfrac{7}{10}\)= 1 . \(\dfrac{6}{5}\)
*Rồi tự làm phần tt đi

Mình ghi kết quả luôn nha bạn