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ĐKXĐ: \(x\ge-2\)
\(\Leftrightarrow x^3+3x\left(x+2\right)-4\left(x+2\right)\sqrt{x+2}=0\)
Đặt \(\sqrt{x+2}=y\ge0\) pt trở thành:
\(x^3+3xy^2-4y^3=0\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+4y^2\right)=0\)
\(\Leftrightarrow x=y\Leftrightarrow\sqrt{x+2}=x\) (\(x\ge0\))
\(\Leftrightarrow x^2=x+2\Leftrightarrow x=2\)
\(ĐKXĐ:x\ge-2\)
\(\Leftrightarrow x^3+3x^2+6x-4x\sqrt{x+2}-8\sqrt{x+2}=0\Leftrightarrow4x^2-4x\sqrt{x+2}+8x-8\sqrt{x+2}+x^3-x\left(x+2\right)=0\Leftrightarrow4x\left(x-\sqrt{x+2}\right)+8\left(x-\sqrt{x+2}\right)+x\left(x-\sqrt{x+2}\right)\left(x+\sqrt{x+2}\right)=0\)\(\Leftrightarrow\left(x-\sqrt{x+2}\right)\left(x^2+x\sqrt{x+2}+4x+8\right)=0\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{x+2}=0\left(1\right)\\x^2+x\sqrt{x+2}+4x+8=0\left(2\right)\end{matrix}\right.\) Từ (1) \(\Rightarrow x=\sqrt{x+2}\left(x\ge0\right)\Rightarrow x^2=x+2\Leftrightarrow x^2-x-2=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\x=-1\left(L\right)\end{matrix}\right.\) Từ (2) \(\Rightarrow x^2+x\sqrt{x+2}+4x+8\ge\left(-2\right)^2+\left(-2\right)\sqrt{-2+2}+4\left(-2\right)+8=4>0\) \(\Rightarrow\) ko có x
vậy...
Bài 5:
a. 1 - 2y + y2
= (1 - y)2
b. (x + 1)2 - 25
= (x + 1)2 - 52
= (x + 1 - 5)(x + 1 + 5)
= (x - 4)(x + 6)
c. 1 - 4x2
= 12 - (2x)2
= (1 - 2x)(1 + 2x)
d. 8 - 27x3
= 23 - (3x)3
= (2 - 3x)(4 + 6x + 9x2)
e. (đề hơi khó hiểu ''x3'' !?)
g. x3 + 8y3
= (x + 2y)(x2 - 2xy + y2)
x2 - 10xy + 9y
= x2 - xy - 9xy + 9y2
= x(x - y) - 9y(x - y)
= (x - y)(x - 9y)
x3 - x2 - 4
= x3 + x2 + 2x - 2x2 - 2x - 4
= x(x2 + x + 2) - 2(x2 + x + 2)
= (x2 + x + 2)(x - 2)
x3 - 5x2 + 8x - 4
= x3 - x2 - 4x2 + 4x + 4x - 4
= x2(x - 1) - 4x(x - 1) + 4(x - 1)
= (x - 1)(x2 - 4x + 4)
= (x - 1)(x - 2)2
x3 + 2x - 3
= x3 - x2 + x2 - x + 3x - 3
= x2(x - 1) + x(x - 1) + 3(x - 1)
= (x - 1)(x2 + x + 3)
x3 + 5x2 + 8x + 4
= x3 + x2 + 4x2 + 4x + 4x + 4
= x2(x + 1) + 4x(x + 1) + 4(x + 1)
= (x + 1)(x2 + 4x + 4)
= (x + 1)(x + 2)2
Em cảm ơn nhiều ạ