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Bài 1 :
\(A=x^2-2xy^2+y^4=\left(x-y^2\right)^2=-\left(y^2-x\right)^2\)
Mà \(B=-\left(y^2-x\right)^2\)
Nên ta có : đpcm
Bài 2
Đặt \(\left(x+1\right)\left(x-2\right)\left(2x-1\right)=0\)
TH1 : x = -1
TH2 : x = 2
TH3 : x = 1/2
Bài 4 :
a, \(\left(2x+3\right)\left(5-x\right)=0\Leftrightarrow x=-\frac{3}{2};5\)
b, \(\left(x-\frac{1}{2}\right)\left(3x+1\right)\left(2-x\right)=0\Leftrightarrow x=\frac{1}{2};-\frac{1}{3};2\)
c, \(x^2+2x=0\Leftrightarrow x\left(x+2\right)=0\Leftrightarrow x=0;-2\)
d, \(x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow x=0;1\)
B1:
Ta có: a - b = ab => a = ab + b = b(a + 1)
Thay a = b(a + 1) vào a - b = a : b ta có: \(a-b=\frac{b\left(a+1\right)}{b}=a+1\)
=> a - b = a + 1 => a - a - b = 1 => -b = 1 => b = -1
Lại có: ab = a - b
<=> a x (-1) = a - (-1) <=> -a = a + 1 <=> -a - a = 1 <=> -2a = 1 <=> a = -1/2
Vậy...
B2:
a, \(3y\left(y-\frac{2}{5}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3y=0\\y-\frac{2}{5}=0\end{cases}\Rightarrow\orbr{\begin{cases}y=0\\y=\frac{2}{5}\end{cases}}}\)
b, \(7\left(y-1\right)+2y\left(y-1\right)=0\)
\(\Rightarrow\left(y-1\right)\left(7+2y\right)=0\)
\(\Rightarrow\orbr{\begin{cases}y-1=0\\7+2y=0\end{cases}\Rightarrow}\orbr{\begin{cases}y=1\\2y=7\end{cases}\Rightarrow}\orbr{\begin{cases}y=1\\y=\frac{7}{2}\end{cases}}\)
B3: \(K=\frac{-2}{3}+\frac{3}{4}-\frac{-1}{6}+\frac{-2}{5}\)
\(K=\left(-\frac{2}{3}+\frac{1}{6}\right)+\left(\frac{3}{4}-\frac{2}{5}\right)\)
\(K=\left(\frac{-4}{6}+\frac{1}{6}\right)+\left(\frac{15}{20}-\frac{8}{20}\right)\)
\(K=\frac{-1}{2}+\frac{7}{20}=\frac{-10}{20}+\frac{7}{20}=\frac{-3}{20}\)
a, => |5/3.x| = 1/6
=> 5/3.x = -1/6 hoặc 5/3.x = 1/6
=> x = -1/10 hoặc x = 1/10
Tk mk nha
a) \(\left|0,5x-2\right|-\left|x+\frac{1}{3}\right|=0\)
=> \(\left|0,5x-2\right|=\left|x+\frac{1}{3}\right|\)
=> \(\orbr{\begin{cases}0,5x-2=x+\frac{1}{3}\\0,5x-2=-x-\frac{1}{3}\end{cases}}\)
=> \(\orbr{\begin{cases}-0,5x=\frac{7}{3}\\1,5x=\frac{5}{3}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{14}{3}\\x=\frac{10}{9}\end{cases}}\)
b) \(2x-\left|x+1\right|=\frac{1}{2}\)
=> \(\left|x+1\right|=2x-\frac{1}{2}\) (Đk: \(2x-\frac{1}{2}\ge0\) <=> \(x\ge\frac{1}{4}\))
=> \(\orbr{\begin{cases}x+1=2x-\frac{1}{2}\\x+1=\frac{1}{2}-2x\end{cases}}\)
=> \(\orbr{\begin{cases}-x=-\frac{3}{2}\\3x=-\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{6}\end{cases}}\)
\(\text{a, }\frac{-2}{5}+x=\left(\frac{-1}{3}\right)^2+\frac{2}{3}\)
\(\Leftrightarrow\frac{-2}{5}+x=\frac{1}{9}+\frac{6}{9}\)
\(\Leftrightarrow\text{ }\frac{-2}{5}+x=\frac{7}{9}\)
\(\Leftrightarrow\text{ }x=\frac{7}{9}-\frac{-2}{5}\)
\(\Leftrightarrow\text{ }x=\frac{53}{45}\)
\(\text{Vậy }x=\frac{53}{45}\)
\(\text{Chia hay cộng mình không biết nên mình làm 2 TH, cái nào đúng đề thì bạn nhìn nha:}\)
\(\text{TH 1: Dấu chia}\)
\(\text{b, }\frac{3}{5}-2x=\left(\frac{-3}{5}\right)^2:\frac{9}{25}\)
\(\text{ }\Leftrightarrow\frac{3}{5}-2x=\frac{9}{25}:\frac{9}{25}\)
\(\text{ }\Leftrightarrow\frac{3}{5}-2x=1\)
\(\text{ }\Leftrightarrow2x=\frac{3}{5}-1\)
\(\text{ }\Leftrightarrow2x=\frac{3}{5}-1\)
\(\text{ }\Leftrightarrow2x=\frac{-2}{5}\)
\(\text{ }\Leftrightarrow x=\frac{-2}{5}:2\)
\(\text{ }\Leftrightarrow x=\frac{-1}{5}\)
\(\text{Vậy }\text{}x=\frac{-1}{5}\)
\(\text{TH 2:Dấu cộng}\)
\(\text{b, }\frac{3}{5}-2x=\left(\frac{-3}{5}\right)^2+\frac{9}{25}\)
\(\Leftrightarrow\frac{3}{5}-2x=\frac{9}{25}+\frac{9}{25}\)
\(\Leftrightarrow\frac{3}{5}-2x=\frac{18}{25}\)
\(\Leftrightarrow2x=\frac{3}{5}-\frac{18}{25}\)
\(\Leftrightarrow2x=\frac{-3}{25}\)
\(\Leftrightarrow x=\frac{-3}{25}:2\)
\(\Leftrightarrow x=\frac{-3}{50}\)
\(\text{Vậy }x=\frac{-3}{50}\)
\(\text{c, }\left|2x-1\right|=\frac{1}{2}-\frac{-2}{3}\)
\(\Leftrightarrow\left|2x-1\right|=\frac{7}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\frac{7}{6}\\2x-1=\frac{-7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\frac{13}{6}\\2x=\frac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{13}{12}\\x=\frac{-1}{12}\end{matrix}\right.\)
\(\text{Vậy }x\in\left\{\frac{13}{12};\frac{-1}{12}\right\}\)
\(\text{d, }\left(x-\frac{3}{4}\right).\frac{1}{2}=\left(\frac{-1}{2}\right)^2\)
\(\Leftrightarrow\left(x-\frac{3}{4}\right).\frac{1}{2}=\frac{1}{4}\)
\(\Leftrightarrow x-\frac{3}{4}=\frac{1}{4}:\frac{1}{2}\)
\(\Leftrightarrow x-\frac{3}{4}=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{2}+\frac{3}{4}\)
\(\Leftrightarrow x=\frac{5}{4}\)
\(\text{Vậy }x=\frac{5}{4}\)
a,
\(\left(\frac{1}{2}\right)^{2x+1}=\frac{1}{32}\)
\(\left(\frac{1}{2}\right)^{2x+1}=\left(\frac{1}{2}\right)^5\)
=>\(2x+1=5\)
2x=5-1
2x=4
x=4:2
x=2
b, mình không biết cách làm
a)\(\left(\frac{1}{2}\right)^{2x+1}=\frac{1}{32}\)
\(\left(\frac{1}{2}\right)^{2x+1}=\left(\frac{1}{2}\right)^5\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow x=2\)