Văn bản ngữ văn 9

cứu với huhu

a: \(A=-x\left(x-y\right)^2+\left(x-y\right)^3+y^3\left(x-2x\right)\)

\(=\left(x-y\right)^2\left(-x+x-y\right)+y^3\cdot\left(-x\right)\)

\(=\left(x-y\right)^2\cdot\left(-y\right)+y^3\cdot\left(-x\right)\)

\(=-y\left(x^2-2xy+y^2\right)-xy^3\)

\(=-x^2y+2xy^2-y^3-xy^3\)

|2x-1|=1

=>\(\left[{}\begin{matrix}2x-1=1\\2x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

Khi x=2 và y=2 thì \(A=-2^2\cdot2+2\cdot2\cdot2^2-2^3-2\cdot2^3\)

=-8+16-8-16

=-16

Khi x=0 và y=2 thì \(A=-0^2\cdot2+2\cdot0\cdot2^2-2^3-0\cdot2^3\)

=-8

b: 

\(\left(x-2\right)^2+y^2=0\)

=>\(\left\{{}\begin{matrix}x-2=0\\y=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)

\(B=-\left(2x-y\right)^3-x\left(2x-y\right)^2-y^3\)

Khi x=2;y=0 thì \(B=-\left(2\cdot2-0\right)^3-2\cdot\left(2\cdot2-0\right)^2-0^3\)

=-64-2*4^2

=-64-32

=-96

c: \(C=\left(x+y\right)\left(x^2-xy+y^2\right)+3\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

\(=x^3+y^3+3\left(8x^3-y^3\right)\)

\(=x^3+y^3+24x^3-3y^3=25x^3-2y^3\)

x+y=2

=>x-3=2

=>x=2+3=5

Khi x=5; y=-3 thì \(C=25\cdot5^3-2\cdot\left(-3\right)^3=3206\)

d: \(D=\left(x+3y\right)\left(x^2-3xy+9y^2\right)+\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)

\(=x^3+27y^3+27x^3-y^3\)

\(=28x^3-26y^3\)

3x-y=5 nên 6-y=5

=>y=1

Khi x=2;y=1 thì \(D=28\cdot2^3-26\cdot1^3=28\cdot8-26=198\)

e: \(E=\left(x+1\right)^3+6\left(x+1\right)^2+12x+20\)

\(=\left(x+1\right)^3+6\left(x+1\right)^2+12\left(x+1\right)+8\)

\(=\left(x+1+2\right)^3=\left(x+3\right)^3\)

Khi x=5 thì \(E=\left(5+3\right)^3=8^3=512\)

f: ta có: \(F=\left(-x-2\right)^3+\left(2x-4\right)\left(x^2+2x+4\right)-x^2\left(x-6\right)\)

\(=-x^3-6x^2-12x-8+2\left(x-2\right)\left(x^2+2x+4\right)-x^3+6x^2\)

\(=-2x^3-12x-8+2\left(x^3-8\right)\)

\(=-2x^3-12x-8+2x^3-16\)

=-12x-24

Khi x=-2 thì \(F=\left(-12\right)\cdot\left(-2\right)-24=24-24=0\)

Bài 123

120.

\(A=\dfrac{x+3a}{2-x}+\dfrac{x-3a}{2+x}-\dfrac{2a}{4-x^2}+a\)

\(=\dfrac{\left(x+3a\right)\left(2+x\right)}{\left(2-x\right)\left(2+x\right)}+\dfrac{\left(x-3a\right)\left(2-x\right)}{\left(2-x\right)\left(2+x\right)}-\dfrac{2a}{\left(2-x\right)\left(2+x\right)}+a\)

\(=\dfrac{x^2+\left(3a+2\right)x+6a-x^2+\left(3a+2\right)x-6a-2a}{\left(2-x\right)\left(2+x\right)}+a\)

\(=\dfrac{2\left(3a+2\right)x-2a}{4-x^2}+a\)

Do \(x=\dfrac{a}{3a+2}\Rightarrow2\left(3a+2\right)x=2a\)

\(\Rightarrow A=\dfrac{2a-2a}{4-x^2}+a=a\) 

121.

Đặt \(\left\{{}\begin{matrix}a-b=x\\b-c=y\\c-a=z\end{matrix}\right.\)

\(\Rightarrow x+y+z=\left(a-b\right)+\left(b-c\right)+\left(c-a\right)=0\)

Ta có:

\(A=\dfrac{2}{x}+\dfrac{2}{y}+\dfrac{2}{z}+\dfrac{x^2+y^2+z^2}{xyz}\)

\(=\dfrac{2\left(xy+yz+zx\right)}{xyz}+\dfrac{x^2+y^2+z^2}{xyz}\)

\(=\dfrac{x^2+y^2+z^2+2\left(xy+yz+zx\right)}{xyz}\)

\(=\dfrac{\left(x+y+z\right)^2}{xyz}=\dfrac{0^2}{xyz}=0\)

_{^{Tham khảo:}}

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