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A= (6x-2)^2 + (2-5x)^2+2(6x-2)(2-5x)
= (6x-2)^2 +2(6x-2)(2-5x)+ (2-5x)^2
\(=\left(6x-2+2-5x\right)^2=x^2\)
B= (2a^2+2a+1)(2a^2-2a+1)-(2a^2+1)^2
\(=\left(2a^2+1\right)^2-4a^2-\left(2a^2+1\right)^2=4a^2\)
C=(x+3)(x^2-3x+9)-(54+x^3)
\(=\left(x^3+27\right)-54-x^3=27\)
D=(2x+y)(4x^2-2xy+y^2)-(2x-y)(4x^2+2xy+y^2)
\(=\left(2x+y\right)^3-\left(2x-y\right)^3\)
E=(a+b)^2-(a-b)^2
\(=\left(a+b+a-b\right)\left(a+b-a+b\right)=2a.2b=4ab\)
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\(C=\left(x+3\right)\left(x^2-3x+9\right)-\left(54+x^3\right)\)
\(=x^3-27-54-x^3=-81\)
\(b,\left(x-1\right)^2-1+x^2=\left(1-x\right)\left(x+3\right)\)
\(x^2-2x+1-1+x^2=x+3-x^2-3x\)
\(2x^2-2x=x+3-x^2-3x\)
\(2x^2-2x=-2x+3-x^2\)
\(2x^2=3-x^2\)
\(2x^2+x^2=3\)
\(3x^2=3\Leftrightarrow x^2=1\Leftrightarrow x=\pm\sqrt{1}\)
tớ n g u nên cần tg suy nghĩ thêm :v
câu a tìm ra r nè , vất vả :v ( kiên trì lắm đấy )
\(a,\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2+1\right)\)
\(9x^3+9x^2-4x-4-3x^2-3x-2x^2-2=0\)
\(6x^3+7x^2-7x-6=0\)
\(\left(6x^2+13x+6\right)\left(x-1\right)=0\)
\(Th1:6x^2+9x+4x+6=0\)
\(\Leftrightarrow\left[3x\left(2x+3\right)+2\left(2x+3\right)\right]=0\)
\(\Leftrightarrow\left(2x+3\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+3=0\\3x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=-3\\3x=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=-\frac{2}{3}\end{cases}}}\)
\(Th2:x-1=0\Leftrightarrow x=1\)
2.Tim x
a,(2x+1)2-4(x+2)2=9
<=> (4x2+4x+1)-4(x2+4x+4)=9
<=> -12x-15=9
<=> -12x=24
<=> x=-2
\(1a,\)\(\left(x^2-0,1\right)=\left(x-\sqrt{0,1}\right)\left(x+\sqrt{0,1}\right)\)
\(1b,\)\(\left(2a^2+b^2\right)^2=\left(2a^2\right)^2+2.2a^2.b^2+\left(b^2\right)^2=4a^4+4a^2b^2+b^4\)
\(1c,\)\(\left(a^2+5\right)\left(5-a^2\right)=\left(5+a^2\right)\left(5-a^2\right)=25-x^4\)
6) c) x3 - x2 + x = 1
<=> x3 - x2 + x - 1 = 0
<=> (x3 - x2) + (x - 1) = 0
<=> x2 (x - 1) + (x - 1) = 0
<=> (x - 1) (x2 + 1) = 0
=> x - 1 = 0 hoặc x2 + 1 = 0
* x - 1 = 0 => x = 1
* x2 + 1 = 0 => x2 = -1 => x = -1
Vậy x = 1 hoặc x = -1
Bài 5:
a) Đặt \(A=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=3^{32}-1\)
\(\Rightarrow A=\frac{3^{32}-1}{8}\)
b) (7x+6)2 + (5-6x)2 - (10-12x)(7x+6)
=(7x+6)2 + (5-6x)2 - 2(5-6x)(7x+6)
\(=\left(7x+6-5+6x\right)^2\)
\(=\left(13x+1\right)^2\)
nhanh giúp mình nhé