Câu hỏi của nam hai

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ai biết làm giúp e vs

dùng hàng đẳng thức vs lm theo cách lớp 8 ạ

Nguyễn Huy Tú · Accepted Answer

Bài 1 $1,\left(x+1\right)^3-\left(x-4\right)\left(x+4\right)-x^3=x^3+3x^2+3x+1-x^2+16-x^3=2x^2+3x+17$2, $\left(x+2\right)^3-x\left(x+3\right)\left(x-3\right)-12x^2-8=x^3+6x^2+12x+8-x\left(x^2-9\right)-12x^2-8$$=x^3+6x^2+12x+8-x^3+9x-12x^2-8=-6x^2+21x$3, $\left(x-2\right)^3-x\left(x-2\right)\left(x+2\right)+6x\left(x-3\right)$$=x^3-6x^2+12x-8-x\left(x^2-4\right)+6x^2-18x$$=x^3+12x-8-x^3+4x-18x=2x=8$4, $x\left(x-5\right)\left(x+5\right)-\left(x-5\right)^3+100x$$=\left(x-5\right)\left[x^2+5x-\left(x-5\right)^2\right]+100x$$=\left(x-5\right)\left(x^2+5x-x^2+10x-25\right)+100x=\left(x-5\right)\left(15x-25\right)+100x$$=15x^2-100x+125+100x=15x^2+125$5, $\left(x-3y\right)^3-\left(x-2y\right)\left(2y+x\right)=x^3-9x^2y+27xy^2-27y^3-x^2+4y^2$6, $\left(-x-2y\right)^3+x\left(2y-x\right)\left(x+2y\right)=-\left(x+2y\right)^3+x\left(2y-x\right)\left(x+2y\right)$$=\left(x+2y\right)\left[-\left(x+2y\right)^2+2xy-x^2\right]=\left(x+2y\right)\left(-x^2-4xy-4y^2+2xy-x^2\right)=\left(x+2y\right)\left(-2x^2-2xy-4y^2\right)$$=\left(x+2y\right)\left(-2x^2-2xy-4y^2\right)=-2x^3-2x^2y-4xy^2-4x^2y+4xy^2-8y^3=-2x^3-6x^2y-8y^3$7, $-\left(2x-y\right)^3-x\left(2x-y\right)^2-y^3$$=-\left(8x^3-12x^2y+6xy^2-y^3\right)-x\left(4x^2-4xy+y^2\right)-y^3=-12x^3+16x^2y-7xy^2$8, $-x\left(x-y\right)^2+\left(x-y\right)^3+y^2\left(y-2x\right)$$=-x\left(x^2-2xy+y^2\right)+x^3-3x^2y+3xy^2-y^3+y^3-2xy^2=-x^2y$Câu trả lời: Bài 1 $1,\left(x+1\right)^3-\left(x-4\right)\left(x+4\right)-x^3=x^3+3x^2+3x+1-x^2+16-x^3=2x^2+3x+17$2, $\left(x+2\right)^3-x\left(x+3\right)\left(x-3\right)-12x^2-8=x^3+6x^2+12x+8-x\left(x^2-9\right)-12x^2-8$$=x^3+6x^2+12x+8-x^3+9x-12x^2-8=-6x^2+21x$3, $\left(x-2\right)^3-x\left(x-2\right)\left(x+2\right)+6x\left(x-3\right)$$=x^3-6x^2+12x-8-x\left(x^2-4\right)+6x^2-18x$$=x^3+12x-8-x^3+4x-18x=2x=8$4, $x\left(x-5\right)\left(x+5\right)-\left(x-5\right)^3+100x$$=\left(x-5\right)\left[x^2+5x-\left(x-5\right)^2\right]+100x$$=\left(x-5\right)\left(x^2+5x-x^2+10x-25\right)+100x=\left(x-5\right)\left(15x-25\right)+100x$$=15x^2-100x+125+100x=15x^2+125$5, $\left(x-3y\right)^3-\left(x-2y\right)\left(2y+x\right)=x^3-9x^2y+27xy^2-27y^3-x^2+4y^2$6, $\left(-x-2y\right)^3+x\left(2y-x\right)\left(x+2y\right)=-\left(x+2y\right)^3+x\left(2y-x\right)\left(x+2y\right)$$=\left(x+2y\right)\left[-\left(x+2y\right)^2+2xy-x^2\right]=\left(x+2y\right)\left(-x^2-4xy-4y^2+2xy-x^2\right)=\left(x+2y\right)\left(-2x^2-2xy-4y^2\right)$$=\left(x+2y\right)\left(-2x^2-2xy-4y^2\right)=-2x^3-2x^2y-4xy^2-4x^2y+4xy^2-8y^3=-2x^3-6x^2y-8y^3$7, $-\left(2x-y\right)^3-x\left(2x-y\right)^2-y^3$$=-\left(8x^3-12x^2y+6xy^2-y^3\right)-x\left(4x^2-4xy+y^2\right)-y^3=-12x^3+16x^2y-7xy^2$8, $-x\left(x-y\right)^2+\left(x-y\right)^3+y^2\left(y-2x\right)$$=-x\left(x^2-2xy+y^2\right)+x^3-3x^2y+3xy^2-y^3+y^3-2xy^2=-x^2y$

HT.Phong (9A5) · Answer

Bài 6:1)$8x^3-12x^2+6x-1=0\
\Leftrightarrow\left(2x-1\right)^3=0\
\Leftrightarrow2x-1=0\
\Leftrightarrow2x=1\
\Leftrightarrow x=\dfrac{1}{2}$2) $x^3-6x^2+12x-8=0\
\Leftrightarrow\left(x-2\right)^3=0\
\Leftrightarrow x-2=0\
\Leftrightarrow x=2$3) $x^2-8x+16=5\left(4-x\right)^3\
\Leftrightarrow\left(x-4\right)^2-5\left(4-x\right)^3=0\
\Leftrightarrow\left(4-x\right)^2-5\left(4-x\right)^3=0\
\Leftrightarrow\left(4-x\right)^2\left[1-5\left(4-x\right)\right]=0\
\Leftrightarrow\left(4-x\right)^2\left(5x-19\right)=0\
\Leftrightarrow\left[{}\begin{matrix}x=4\x=\dfrac{19}{5}\end{matrix}\right.$4) $\left(2-x\right)^3=6x\left(x-2\right)\
\Leftrightarrow\left(2-x\right)^3-6x\left(x-2\right)=0\
\Leftrightarrow\left(2-x\right)^3+6x\left(2-x\right)=0\
\Leftrightarrow\left(2-x\right)\left[\left(2-x\right)^2+6x\right]=0\
\Leftrightarrow\left(2-x\right)\left(4-4x+x^2+6x\right)=0\
\Leftrightarrow\left(2-x\right)\left(x^2+2x+4\right)=0\
\Leftrightarrow2-x=0\
\Leftrightarrow x=2$(vì x^2+2x+4=x^2+2x+1+3=(x+1)^2+3>0) Câu trả lời: Bài 6:1)$8x^3-12x^2+6x-1=0\
\Leftrightarrow\left(2x-1\right)^3=0\
\Leftrightarrow2x-1=0\
\Leftrightarrow2x=1\
\Leftrightarrow x=\dfrac{1}{2}$2) $x^3-6x^2+12x-8=0\
\Leftrightarrow\left(x-2\right)^3=0\
\Leftrightarrow x-2=0\
\Leftrightarrow x=2$3) $x^2-8x+16=5\left(4-x\right)^3\
\Leftrightarrow\left(x-4\right)^2-5\left(4-x\right)^3=0\
\Leftrightarrow\left(4-x\right)^2-5\left(4-x\right)^3=0\
\Leftrightarrow\left(4-x\right)^2\left[1-5\left(4-x\right)\right]=0\
\Leftrightarrow\left(4-x\right)^2\left(5x-19\right)=0\
\Leftrightarrow\left[{}\begin{matrix}x=4\x=\dfrac{19}{5}\end{matrix}\right.$4) $\left(2-x\right)^3=6x\left(x-2\right)\
\Leftrightarrow\left(2-x\right)^3-6x\left(x-2\right)=0\
\Leftrightarrow\left(2-x\right)^3+6x\left(2-x\right)=0\
\Leftrightarrow\left(2-x\right)\left[\left(2-x\right)^2+6x\right]=0\
\Leftrightarrow\left(2-x\right)\left(4-4x+x^2+6x\right)=0\
\Leftrightarrow\left(2-x\right)\left(x^2+2x+4\right)=0\
\Leftrightarrow2-x=0\
\Leftrightarrow x=2$(vì x^2+2x+4=x^2+2x+1+3=(x+1)^2+3>0)

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