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Mik cần lời giải á, các bạn toàn cho mik đáp án hoặc là cho mỗi câu 123 (Q▪︎Q)
1: =(x+y)^2+x+y
=(x+y)(x+y+1)
2: =>3x-9+18<4x+2
=>4x+2>3x+9
=>x>7
-x2-8x-16=0
\(\Leftrightarrow-\left(x^2+8x+16\right)=0\)
\(\Leftrightarrow\left(x+4\right)^2=0\)
\(\Leftrightarrow x+4=0\Leftrightarrow x=-4\)
-x2-8x-16=0
\(\Leftrightarrow x^2+8+16=0\)( vì mình chuyển vế nhé )
\(\Leftrightarrow\left(x+4\right)^2=0\)
\(\Leftrightarrow x=-4\)
vậy x=-4
Bài 3:
a) \(\left(2-3x\right)^2-\left(3-x\right)^2=\left[\left(2-3x\right)-\left(3-x\right)\right]\left[\left(2-3x\right)+\left(3-x\right)\right]\)
\(=\left(-1-2x\right)\left(5-4x\right)\)
b) \(49\left(x-3\right)^2-9\left(x+2\right)^2\)
\(=\left[7\left(x-3\right)\right]^2-\left[3\left(x+2\right)\right]^2\)
\(=\left[\left(7x-21\right)-\left(3x+6\right)\right]\left[\left(7x-21\right)+\left(3x+6\right)\right]\)
\(=\left(4x-27\right)\left(10x-15\right)\)
c) \(2xy-x^2-y^2+16=16-\left(x-y\right)^2=\left(16-x+y\right)\left(16+x-y\right)\)
d) \(2\left(x-3\right)+3\left(x^2-9\right)=2\left(x-3\right)+3\left(x-3\right)\left(x+3\right)\)
\(=\left(x-3\right)\left(3x+11\right)\)
e) \(16x^2-\left(x^2+4\right)^2=\left(4x-x^2-4\right)\left(4x+x^2+4\right)\)
\(=-\left(x-2\right)^2\left(x+2\right)^2\)
f) \(1-2x+2yz+x^2-y^2-z^2=\left(x-1\right)^2-\left(y-z\right)^2\)
\(=\left(x-1-y+z\right)\left(x-1+y-z\right)\)
\(C=6\left(c-d\right)\left(c+d\right)\left(c+d\right)+12\left(c-d\right)\left(c-d\right)\left(c+d\right)+c^3+3c^2d+3cd^2+d^3+8\left(c^3-3c^2d+3cd^2-d^3\right)\)
\(C=6\left(c^2-d^2\right)\left(c+d\right)+12\left(c-d\right)\left(c^2-d^2\right)+c^3+3c^2d+3cd^2+d^3+8\left(c^3-3c^2d+3cd^2-d^3\right)\)\(C=6\left(c^3+c^2d-cd^2-d^3\right)+12\left(c^3-c^2d-cd^2-d^2\right)+c^3+3c^2d+3cd^2+d^3+8\left(c^3-3c^2d+3cd^2-d^3\right)\)
\(C=27c^3-27c^2d-39cd^2-25d^3\)