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\(\left(\frac{x-1}{x+2}\right)^2-4\left(\frac{x^2-1}{x^2-4}\right)^2+3\left(\frac{x+1}{x-2}\right)^2=0\left(1\right)\)
\(ĐKXĐ:x\ne\pm2\)
Đặt \(\frac{x-1}{x+2}=a;\frac{x+1}{x-2}=b\)
=> Phương trình (1) <=> \(a^2-4ab+3b^2=0\)
\(\Leftrightarrow a^2-3ab-ab+3b^2=0\)
\(\Leftrightarrow a\left(a-b\right)-3b\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-3b\right)\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-3b\right)\left(a-b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a-3b=0\\a-b=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=3b\\a=b\end{cases}}}\)
=> \(b=0;a=0\)
Bạn cùng trường :">
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: \(\left(x-3\right)^2=\left(-2\right)^2\)
\(\Rightarrow x-3=-2\)
\(\Rightarrow x=-2+3\)
\(\Rightarrow x=1\)
\(\Leftrightarrow x\in\left\{1;5\right\}\)
\(A=x^2+2xy+y^2-4x-4y+q\)
\(=\left(x+y\right)^2-4\left(x+y\right)+q\)
\(=3^2-4.3+q\)
\(=q-3\)
(x-3)(x2 + 3x +9)- ((x-4)((x+4)=21
x3 - 27 - x2 + 4 = 21
x2 + x - 27 -x2 + 4 =21
x=27 -4 + 21
x= 44
2)
( x _ 3 ) ( x^2 + 3x + 9 ) - ( x - 4 ) . ( x + 4 ) = 21
= x^3 - 9 - x^2 - 2^2 = 21
= x - 9 - x^2 - 4
= x^3 - x^2 - 9 - 4
x = - 9 - 4 = - 13
(X+2)-(x-3)=7x-2(x+1)
2x-1=5x-2
2x-1+1=5x-2+1
2x=5x-1
2x-5x=5x-1-5x
-3x=1
-3x/-3=-1/-3
X=1/3
Vậy x=1/3