\(x^2=900\Leftrightarrow x^2=30^2\Rightarrow x=30\)

Chọn A

mik nghĩ A

a) \(\frac{x-1}{-15}=\frac{-60}{x-1}\)

\(\left(x-1\right)^2=\left(-15\right).\left(-60\right)=900\)

\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^2=300^2\\\left(x-1\right)^2=\left(-300\right)^2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x-1=300\\x-1=-300\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=301\\x=-299\end{cases}}\)

b) \(\left|x+\frac{4}{5}\right|+\frac{3}{5}=\frac{2}{5}\)

\(\left|x+\frac{4}{5}\right|=\frac{2}{5}-\frac{3}{5}\)

\(\left|x+\frac{4}{5}\right|=\frac{-1}{5}\)

vì \(\left|x+\frac{4}{5}\right|\ge0\forall x\)mà \(\left|x+\frac{4}{5}\right|=\frac{-1}{5}\)

\(\Rightarrow\)không có giá trị x nào thỏa mãn đề bài trên

c) \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)

\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)

\(\Rightarrow x-\frac{1}{2}=\frac{1}{3}\)

\(\Rightarrow x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)

a) \(\Leftrightarrow\left(x-1\right)\left(x-1\right)=\left(-60\right).\left(-15\right)\)

\(\Leftrightarrow\left(x-1\right)^2=900=30^2\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=30\\x-1=-30\end{cases}\Leftrightarrow\orbr{\begin{cases}x=30+1\\x=-30+1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=31\\x=-29\end{cases}}}\)

Vậy x = 31 hoặc x = - 29

b) \(\left|x+\frac{4}{5}\right|+\frac{3}{5}=\frac{2}{5}\)

\(\Leftrightarrow\left|x+\frac{4}{5}\right|=\frac{2}{5}-\frac{3}{5}\)

\(\Leftrightarrow\left|x+\frac{4}{5}\right|=\frac{-1}{5}\)vô lý không có giá trị tuyệt đối của số nào mà nhận giá trị âm

Vậy ko có giá trị nào của x thỏa mãn

c) \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)

\(\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}\)

\(\Leftrightarrow x=\frac{1}{3}+\frac{1}{2}\)

\(\Leftrightarrow x=\frac{5}{6}\)

a) Ta có: \(\dfrac{x}{-15}=\dfrac{-60}{x}\)

\(\Leftrightarrow x^2=\left(-15\right)\cdot\left(-60\right)=900\)

hay \(x\in\left\{30;-30\right\}\)

Vậy: \(x\in\left\{30;-30\right\}\)

b) Ta có: \(\left|x\right|+0.573=2\)

\(\Leftrightarrow\left|x\right|=1.427\)

hay \(x\in\left\{1.427;-1.427\right\}\)

Vậy: \(x\in\left\{1.427;-1.427\right\}\)

c) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=-1\)

\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{8}{3};-\dfrac{10}{3}\right\}\)

d) Ta có: \(0.01:2.5=\left(0.75x\right):0.75\)

\(\Leftrightarrow\dfrac{0.75\cdot x}{0.75}=\dfrac{0.01}{2.5}\)

\(\Leftrightarrow x=\dfrac{1}{250}\)

Vậy: \(x=\dfrac{1}{250}\)

Tìm x trong tỉ lệ thức

a)

a) x vô nghĩa

b) x=0,8;x=-0,8

d: Đặt x/2=y/5=k

=>x=2k; y=5k

Ta có: xy=10

nên k²=1

Trường hợp 1: k=1

=>x=2; y=5

Trường hợp 2: k=-1

=>x=-2; y=-5

Bài 2: 

a: Ta có: \(\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{60}+6\right):2\sqrt{3}\)

\(=\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{12}\left(\sqrt{5}+\sqrt{3}\right):2\sqrt{3}\)

\(=2\sqrt{12}:2\sqrt{3}\)

=2

b: Ta có: \(\sqrt{5-\sqrt{21}}-\sqrt{\dfrac{7}{2}}\)

\(=\dfrac{\sqrt{10-2\sqrt{21}}-\sqrt{7}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{7}-\sqrt{3}-\sqrt{7}}{\sqrt{2}}\)

\(=-\dfrac{\sqrt{6}}{2}\)

 60x [7/12+4/15]

60x153/180

=9180/180

b 1/2x2/3x3/4x4/5x5/6x6/7x7/8x8/9=40320/4032

\(\frac{x-1}{15}=\frac{-60}{1-x}\)

\(\Rightarrow\frac{x-1}{15}=\frac{60}{-\left(1-x\right)}\)

\(\Rightarrow\frac{x-1}{15}=\frac{60}{x-1}\)

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

\(\Rightarrow\left[\begin{matrix}x-1=30\\x-1=-30\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=31\\x=-29\end{matrix}\right.\)

Vậy \(x\in\left\{31;-29\right\}\)

\(\frac{x-1}{15}=\frac{-60}{1-x}\)

\(\Rightarrow\frac{x-1}{15}=\frac{-60}{-x+1}\)

\(\Rightarrow\frac{x-1}{15}=\frac{-60}{-\left(x-1\right)}\)

\(\Rightarrow\frac{x-1}{15}=\frac{60}{x-1}\)

\(\Rightarrow\left(x-1\right)^2=15.60\)

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

\(\Rightarrow x-1=\pm30\)

\(\Rightarrow\left\{\begin{matrix}x-1=30\\x-1=-30\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=31\\x=-29\end{matrix}\right.\)

Vậy...

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