Giải:

\(x-5\sqrt{x}\) = 0 (\(x\) ≥ 0)

\(\sqrt{x}\) .(\(\sqrt{x}\) - 5) = 0

\(\left[\begin{array}{l}\sqrt{x}=0\\ \sqrt{x}-5=0\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ \sqrt{x}=5\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x=25\end{array}\right.\)

Vậy \(x\in\) {0; 25}

\(x^5\) = 2\(x^7\)

\(x^5\) - 2\(x^7\) = 0

\(x^5\).(1 - 2\(x^2\)) = 0

\(\left[\begin{array}{l}x^5=0\\ 1-2x^2=0\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ 2x^2=1\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x^2=\frac12\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x=\pm\sqrt{\frac12}\end{array}\right.\)

Vậy \(x\) ∈ {- \(\sqrt{\frac12}\); 0; \(\sqrt{\frac12}\)}

Giải:

\(x-5\sqrt{x}\) = 0 (\(x\) ≥ 0)

\(\sqrt{x}\) .(\(\sqrt{x}\) - 5) = 0

\(\left[\begin{array}{l}\sqrt{x}=0\\ \sqrt{x}-5=0\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ \sqrt{x}=5\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x=25\end{array}\right.\)

Vậy \(x\in\) {0; 25}

\(x^5\) = 2\(x^7\)

\(x^5\) - 2\(x^7\) = 0

\(x^5\).(1 - 2\(x^2\)) = 0

\(\left[\begin{array}{l}x^5=0\\ 1-2x^2=0\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ 2x^2=1\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x^2=\frac12\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x=-\frac{1}{\sqrt2}\\ x=\frac{1}{\sqrt2}\end{array}\right.\)

Vậy \(x\) \(\in\) {- \(\frac{1}{\sqrt2}\); 0; \(\frac{1}{\sqrt2}\)}

\(\left|x\right|=7\)

\(\Rightarrow\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)

Vậy \(x\in\left\{\pm7\right\}\)

\(\left|x\right|=0\)

\(\Rightarrow x=0\)

Vậy x = 0

a)Ta có: \(14x=12y\Rightarrow\frac{x}{12}=\frac{y}{14}=\frac{x-y}{12-14}=\frac{-10,2}{-2}=5,1\)

\(\Rightarrow x=5,1.12=61,2\)

     \(y=5,1.14=71,4\)

b) Ta có: \(\left(x-5\right)^{2016}-\left|y^2-4\right|=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x-5\right)^{2016}=0\\y^2-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x-5=0\\y^2=4\end{cases}\Rightarrow}\orbr{\begin{cases}x=5\\y=\pm2\end{cases}}}\)

Vậy....

1) \(\left|x\right|=7\)

=> \(\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\)

Vậy \(x\in\left\{7;-7\right\}.\)

2) \(\left|x\right|=0\)

=> \(x=0\)

Vậy \(x\in\left\{0\right\}.\)

5) \(\left|x\right|-1=\frac{2}{5}\)

=> \(\left|x\right|=\frac{2}{5}+1\)

=> \(\left|x\right|=\frac{7}{5}\)

=> \(\left[{}\begin{matrix}x=\frac{7}{5}\\x=-\frac{7}{5}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{7}{5};-\frac{7}{5}\right\}.\)

8) \(\left|x-17\right|=23\)

=> \(\left[{}\begin{matrix}x-17=23\\x-17=-23\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=23+17\\x=\left(-23\right)+17\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=40\\x=-6\end{matrix}\right.\)

Vậy \(x\in\left\{40;-6\right\}.\)

Mình chỉ làm thế thôi nhé, bạn đăng hơi nhiều mà với cả mấy câu này dễ mà bạn.

Chúc bạn học tốt!

1)

=>

Vậy

2)

=>

Vậy

5)

=>

=>

=>

Vậy

8)

=> =>

=>

Vậy

 a)     \(\left(2x-3\right)\left(\frac{3}{4}x+1\right)=0\)

<=>\(\hept{\begin{cases}2x-3=0\\\frac{3}{4}x+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x=3\\\frac{3}{4}x=-1\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\x=-\frac{3}{4}\end{cases}}}\)

b)        \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}\Leftrightarrow\hept{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}}\)

Giải:

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

\(\Leftrightarrow x.x=-4.\left(-9\right)\)

\(\Leftrightarrow x^2=36\)

\(\Leftrightarrow x=\pm6\)

Vậy ...

b) \(\dfrac{x-1}{-15}=\dfrac{-60}{x-1}\)

\(\Leftrightarrow\left(x-1\right)^2=900\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=30\\x-1=-30\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=31\\x=-29\end{matrix}\right.\)

Vậy ...

d) \(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\)

\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=\left(x-1\right)\left(x+4\right)\)

\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)

\(\Leftrightarrow5x-14=3x-4\)

\(\Leftrightarrow2x=10\)

\(\Leftrightarrow x=5\)

Vậy ...

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{x}{y+z-5}=\frac{y}{x+z+3}=\frac{z}{x+y+2}=\frac{x+y+z}{y+z-5+x+z+3+x+y+2}=\frac{x+y+z}{2x+2y+2z}=\frac12\)

=>\(\begin{cases}y+z-5=2x\\ x+z+3=2y\\ x+y+2=2z\end{cases}\Rightarrow\begin{cases}y+z=2x+5\\ y+z=2y-3\\ x+y=2z-2\end{cases}\)

\(\frac{x}{y+z-5}=\frac12\left(x+y+z\right)\)

=>\(\frac12\left(x+y+z\right)=\frac12\)

=>x+y+z=1

*Ta có: x+y+z=1

=>z+2z-2=1

=>3z-2=1

=>3z=3

=>z=1

*Ta có: x+y+z=1

=>y+2y-3=1

=>3y=4

=>\(y=\frac43\)

*Ta có: x+y+z=1

=>x+2x+5=1

=>3x+5=1

=>3x=-4

=>\(x=-\frac43\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{x}{y+z-5}=\frac{y}{x+z+3}=\frac{z}{x+y+2}=\frac{x+y+z}{y+z-5+x+z+3+x+y+2}=\frac{x+y+z}{2x+2y+2z}=\frac12\)

=>\(\begin{cases}y+z-5=2x\\ x+z+3=2y\\ x+y+2=2z\end{cases}\Rightarrow\begin{cases}y+z=2x+5\\ y+z=2y-3\\ x+y=2z-2\end{cases}\)

\(\frac{x}{y+z-5}=\frac12\left(x+y+z\right)\)

=>\(\frac12\left(x+y+z\right)=\frac12\)

=>x+y+z=1

*Ta có: x+y+z=1

=>z+2z-2=1

=>3z-2=1

=>3z=3

=>z=1

*Ta có: x+y+z=1

=>y+2y-3=1

=>3y=4

=>\(y=\frac43\)

*Ta có: x+y+z=1

=>x+2x+5=1

=>3x+5=1

=>3x=-4

=>\(x=-\frac43\)

c) Bạn dùng "tích trung tỉ bằng tích ngoại tỉ"

d) \(\frac{x-1}{4}=\frac{9}{x-1}\)

=> (x - 1)² = 4 . 9 = 36 = (+ 6)²

=> x - 1 = 6 hoặc x - 1 = -6

=> x = 7 hoặc x = -5