(x+1/2)2 =1/6
Giúp mình cái bài này nhanh nha vì mình sắp phải ik học òi:3
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NL
Nguyễn Lê Phước Thịnh
CTVHS
16 tháng 10 2025
1: \(\frac{2x+6}{3x^2-x}:\frac{x^2+3x}{1-3x}\)
\(=\frac{2\left(x+3\right)}{x\left(3x-1\right)}\cdot\frac{-3x+1}{x\left(x+3\right)}\)
\(=\frac{2}{x}\cdot\frac{-\left(3x-1\right)}{x\left(3x-1\right)}=\frac{-2}{x^2}\)
2: \(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}\)
\(=\frac{x}{x-2y}+\frac{x}{x+2y}-\frac{4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{x\left(x+2y\right)+x\left(x-2y\right)-4xy}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x^2-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x}{x+2y}\)
3: \(\frac{1}{3x-2}-\frac{1}{3x+2}-\frac{3x-6}{4-9x^2}\)
\(=\frac{1}{3x-2}-\frac{1}{3x+2}+\frac{3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\frac{3x+2-\left(3x-2\right)+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\frac{3x+2-3x+2+3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\frac{3x-2}{\left(3x-2\right)\left(3x+2\right)}=\frac{1}{3x+2}\)
4: \(\frac{x+3}{x+1}+\frac{2x-1}{x-1}+\frac{x+5}{x^2-1}\)
\(=\frac{x+3}{x+1}+\frac{2x-1}{x-1}+\frac{x+5}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{\left(x+3\right)\left(x-1\right)+\left(2x-1\right)\left(x+1\right)+x+5}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+2x-3+2x^2+2x-x-1+x+5}{\left(x-1\right)\left(x+1\right)}=\frac{3x^2+4x+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{3x+1}{x-1}\)
11 tháng 2 2018
\(\left(x+1\right)\left(y-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\y-2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0-1\\y=0+2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-1\\y=2\end{cases}}\)
Vậy x = -1 hoặc y = 2
LN
1
4 tháng 7 2016
ta xét :\(2^4-4^2=2^4-\left(2^2\right)^2=2^4-2^4=0\)
=> giá trị biểu thức =0
BB
1
26 tháng 2 2022
a, \(\dfrac{4x}{7}=\dfrac{1}{5}+\dfrac{2}{3}=\dfrac{3+10}{15}=\dfrac{13}{15}\Rightarrow60x=91\Leftrightarrow x=\dfrac{91}{60}\)
b, \(\dfrac{5}{7}:x=\dfrac{1}{6}+\dfrac{4}{5}=\dfrac{5+24}{30}=\dfrac{29}{30}\Leftrightarrow x=\dfrac{5}{7}:\dfrac{29}{30}=\dfrac{150}{203}\)
có thể bn vt sai đề nhưng mk làm theo cái đề nha
\(\left(x+\frac{1}{2}\right)^2=\left(\sqrt{\frac{1}{6}}\right)^2=\left(-\sqrt{\frac{1}{6}}\right)^2\)
\(=>\orbr{\begin{cases}x=\sqrt{\frac{1}{6}}-\frac{1}{2}\\x=-\sqrt{\frac{1}{6}}-\frac{1}{2}\end{cases}}\)
Vậy .....
\(\left(\left(x+\frac{1}{2}\right)\cdot2\right)-\frac{1}{6}=0\)
\(\frac{x\cdot2+1}{2}=\frac{2x+1}{2}\)
\(\left(\frac{\left(x\cdot2+1\right)}{2}\cdot2\right)-\frac{1}{6}=0\)
\(\left(2x+1\right)-\frac{1}{6}=0\)
\(2x+1=\frac{2x+1}{1}=\frac{\left(2x+1\right)\cdot6}{6}\)
\(\frac{\left(2x+1\right)\cdot6-\left(1\right)}{6}=\frac{12x+5}{6}\)
\(\frac{12x+5}{6}\cdot6=0\cdot6\)
\(12x+5=0\)
\(x=\frac{-5}{12}=0.417\)