Văn bản ngữ văn 9

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

câu hỏi đâu nhỉ em?

K có câu hỏi sao​

câu hỏi đâu bn???

cứu với huhu

Bài 115

Bài 117

113.

\(\dfrac{x^2}{2}+\dfrac{y^2}{3}+\dfrac{z^2}{4}=\dfrac{x^2+y^2+z^2}{5}\)

\(\Leftrightarrow\dfrac{x^2}{2}-\dfrac{x^2}{5}+\dfrac{y^2}{3}-\dfrac{y^2}{5}+\dfrac{z^2}{4}-\dfrac{z^2}{5}=0\)

\(\Leftrightarrow\dfrac{3}{10}x^2+\dfrac{2}{15}y^2+\dfrac{1}{20}z^2=0\)

Do \(x^2;y^2;z^2\ge0;\forall x;y;z\)

\(\Rightarrow\dfrac{3}{10}x^2+\dfrac{2}{15}y^2+\dfrac{1}{20}z^2\ge0;\forall x;y;z\)

Dấu "=" xảy ra khi và chỉ khi \(x=y=z=0\)

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:

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

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

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

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

Nhiều bài quá e, nên làm những bài dễ rùi bài khó hãng hỏi nhé

150.

\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\Rightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)

\(\Rightarrow ayz+bxz+cxy=0\)

\(x+y+z=0\Rightarrow\left\{{}\begin{matrix}-x=y+z\\-y=x+z\\-z=x+y\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=\left(y+z\right)^2\\y^2=\left(x+z\right)^2\\z^2=\left(x+y\right)^2\end{matrix}\right.\)

Do đó:

\(ax^2+by^2+cz^2=a\left(y+z\right)^2+b\left(x+z\right)^2+c\left(x+y\right)^2\)

\(=a\left(y^2+z^2+2yz\right)+b\left(x^2+z^2+2xz\right)+c\left(x^2+y^2+2xy\right)\)

\(=x^2\left(b+c\right)+y^2\left(a+c\right)+z^2\left(a+b\right)+2\left(ayz+bxz+cxy\right)\)

\(=x^2.\left(-a\right)+y^2\left(-b\right)+z^2.\left(-c\right)+2.0\)

\(=-\left(ax^2+by^2+cz^2\right)\)

\(\Rightarrow2\left(ax^2+by^2+cz^2\right)=0\)

\(\Rightarrow ax^2+by^2+cz^2=0\)

151.

Biểu thức xác định khi \(xyz\ne0\)

Giả thiết tương đương:

\(x+\dfrac{1}{y}=y+\dfrac{1}{z}=z+\dfrac{1}{x}\)

Từ \(x+\dfrac{1}{y}=y+\dfrac{1}{z}\Rightarrow x-y=\dfrac{1}{z}-\dfrac{1}{y}\)

\(\Rightarrow yz\left(x-y\right)=y-z\) (1)

Tương tự: \(y+\dfrac{1}{z}=z+\dfrac{1}{x}\Rightarrow xz\left(y-z\right)=z-x\) (2)

\(x+\dfrac{1}{y}=z+\dfrac{1}{x}\Rightarrow xy\left(z-x\right)=x-y\) (3)

- Nếu tồn tại 2 trong 3 số x;y;z bằng nhau, giả sử \(x=y\)

Thay vào (3) \(\Rightarrow xy\left(z-x\right)=0\Rightarrow z=x\)

\(\Rightarrow x=y=z\)

- Nếu x;y;z đôi một phân biệt \(\Rightarrow\left(x-y\right)\left(y-z\right)\left(z-x\right)\ne0\)

Nhân vế (1),(2) và (3):

\(\Rightarrow x^2y^2z^2\left(x-y\right)\left(y-z\right)\left(z-x\right)=\left(x-y\right)\left(y-z\right)\left(z-x\right)\)

\(\Rightarrow x^2y^2z^2=1\)

142.

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{a}\)

\(\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\)

\(\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-\dfrac{1}{x+y+z}=0\)

\(\Rightarrow\dfrac{x+y}{xy}+\dfrac{x+y}{z\left(x+y+z\right)}=0\)

\(\Rightarrow\left(x+y\right)\left(\dfrac{1}{xy}+\dfrac{1}{z\left(x+y+z\right)}\right)=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\z+x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x+y+z=z\\x+y+z=x\\x+y+z=y\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a=z\\a=x\\a=y\end{matrix}\right.\)

143.

Biểu thức xác định khi \(xyz\ne0\)

Giả sử \(x+y+z\) và \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\) đồng thời bằng 0

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)

\(\Rightarrow\dfrac{xy+yz+zx}{xyz}=0\)

\(\Rightarrow xy+yz+zx=0\) (1)

Lại có:

\(x+y+z=0\)

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

\(\Rightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=0\) (2)

\(\left(1\right);\left(2\right)\Rightarrow x^2+y^2+z^2=0\)

\(\Rightarrow x=y=z=0\) (ko thỏa mãn ĐKXĐ)

Vậy điều giả sử là sai hay \(x+y+z\) và \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\) ko thể đồng thời bằng 0

144.

\(\left\{{}\begin{matrix}2a=by+cz\left(1\right)\\2b=ax+cz\left(2\right)\\2c=ax+by\left(3\right)\end{matrix}\right.\)

Cộng vế (1), (2) và (3):

\(\Rightarrow a+b+c=ax+by+cz\) (4)

Lần lượt trừ vế (4) cho (1), (2) và (3):

\(\Rightarrow\left\{{}\begin{matrix}ax=b+c-a\\by=a+c-b\\cz=a+b-c\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{b+c-a}{a}\\y=\dfrac{a+c-b}{b}\\z=\dfrac{a+b-c}{c}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+2=\dfrac{a+b+c}{a}\\y+2=\dfrac{a+b+c}{b}\\z+2=\dfrac{a+b+c}{c}\end{matrix}\right.\)

\(\Rightarrow M=\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}=1\)