Để \(x\left(2+x^2\right)\left(7-x\right)=0\hept{\begin{cases}x=0\\2+x^2=0\\7-x=0\end{cases}}\)

Mà \(2+x^2>0\) => \(\orbr{\begin{cases}x=0\\7-x=0\end{cases}}\)=> \(\orbr{\begin{cases}x=0\\x=7\end{cases}}\)

Vậy x = { 0; 7 }

bn có chắc ko,dù sao cũng cảm ơn bn

b) x^2 - 3x = 0

<=> x(x - 3) = 0

<=> x = 0 hoặc x - 3 = 0

                        x = 3

=> x = 0 hoặc x = 3

Chihiro CTV làm hết đi nhe

\(\left(x-1\right)^2-x\left(5-x\right)=0\)

\(\Leftrightarrow x^2-2x+1-5x+x^2=0\)

\(\Leftrightarrow2x^2-7x+1=0\)

Ta có: \(\Delta=7^2-4.8=25,\sqrt{\Delta}=5\)

\(\orbr{\begin{cases}x_1=\frac{7+5}{4}=3\\x_2=\frac{7-5}{4}=\frac{1}{2}\end{cases}}\)

a) \(\left(x-3\right)^2-4=0\)

\(\left(x-7\right)\left(x+1\right)=0\)

\(\orbr{\begin{cases}x=7\\x=-1\end{cases}}\)

b) \(x^2-2x=24\)

\(x^2-2x-24=0\)

\(\left(x-6\right)\left(x+4\right)=0\)

\(\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)

c) \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(4x^2+4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

\(5x^2+10x+10-5x^2+245=0\)

\(10x+255=0\)

\(x=-25.5\)

A) \(\left(x-3\right)^2-4=0\)

\(\left(x-3\right)^2=4\Rightarrow\left(x-3\right)^2=\left(-2\right)^2;2^2\)

th1\(\left(x-3\right)^2=2^2\)

\(\Rightarrow x-3=2\)

\(\Rightarrow x=2+3\)

\(\Rightarrow x=5\)

th2: \(\left(x-3\right)^2=\left(-2\right)^2\)

\(\Rightarrow x-3=-2\)

\(\Rightarrow x=-2+3\)

\(\Rightarrow x=1\)

\(\Leftrightarrow x\in\left\{1;5\right\}\)

a) \(\left(x-3\right)^2-4=0\)

Ta xét 2TH:

TH1: \(x-3=2\)

\(x-3+3=2+3\)

\(x=5\)

TH2: \(x-3=-2\)

\(x-3+3=-2+3\)

\(x=1\)

=> \(x\in\left\{1;5\right\}\)

cmt gi may co bai tu di ma lam bat ng ta lam cho may chep ha

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

<=>\(2x=\frac{-2}{3}-\frac{-1}{2}=\frac{-1}{6}\)

<=>\(x=\frac{-1}{6}:2=-\frac{1}{2}\)

Vây \(x=-\frac{1}{2}\)

\(0.75-\left(-2x\right)=\frac{4}{5}\)

<=>\(\frac{3}{5}+2x=\frac{4}{5}\)

<=>\(2x=\frac{4}{5}-\frac{3}{5}=\frac{1}{5}\)

<=>\(x=\frac{1}{5}:2=\frac{1}{10}\)

Vậy \(x=\frac{1}{10}\)

\(\left(2x+5\right)\left(1-x\right)=0\)

<=>\(2x+5=0\)hoặc \(1-x=0\)

<=>\(x=\frac{5}{2}\)hoặc \(x=1\)

Vậy \(x=\frac{5}{2}\)hoặc \(x=1\)

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

\(\Rightarrow2x-\frac{1}{2}=\frac{-2}{3}\)

\(\Rightarrow2x=\frac{-2}{3}+\frac{1}{2}\)

\(\Rightarrow2x=\frac{-1}{6}\)

\(\Rightarrow x=\frac{-1}{6}:2\)

\(\Rightarrow x=\frac{-1}{12}\)

Vậy  \(x=\frac{-1}{12}\)

b) \(0,75-\left(-2x\right)=\frac{4}{5}\)

\(\Rightarrow\frac{3}{4}+2x=\frac{4}{5}\)

\(\Rightarrow2x=\frac{4}{5}-\frac{3}{4}\)

\(\Rightarrow2x=\frac{1}{20}\)

\(\Rightarrow x=\frac{1}{20}:2\)

\(\Rightarrow x=\frac{1}{40}\)

Vậy  \(x=\frac{1}{40}\)

c) \(\left(2x+5\right)\left(1-x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+5=0\\1-x=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=-5\\x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-5}{2}\\x=1\end{cases}}\)

Vậy  \(x=\frac{-5}{2}\)hoặc \(x=1\)

_Chúc bạn học tốt_

A = |3,7 - x| + 2,5 

=> |3,7 - x| = -2.5

=> ko có giá trị nào cũng x thỏa mã đề bài

 B = |x + 1,5| - 4,5  

=> |x + 1,5| = 4.5

1) x+1.5 = 4.5

x= 3 

2) x+1.5 = -4.5 

x=-4.5 - 1.5

x= -5.5

= 1,736

a,  (x-5)x=0 

->x=0 và x-5=0 

->x=0 và x=5

\(\left|x+3\right|+\left|x-2\right|=5\left(1\right)\)

Xét \(x\le-3\), \(\left(1\right)\Leftrightarrow-x-3+2-x=5\)

\(\Leftrightarrow-2x=6\Leftrightarrow x=-3\) (thỏa mãn)

Xét \(-3< x\le2\), \(\left(1\right)\Leftrightarrow x+3+2-x=5\)

\(\Leftrightarrow5=5\Leftrightarrow x\in Z\)

Xét \(x>5\), \(\left(1\right)\Leftrightarrow x+3+x-2=5\)

\(\Leftrightarrow2x=4\Leftrightarrow x=2\) (loại)