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ĐKXĐ: \(x\ne\left\{-\frac{1}{2};\frac{1}{2};-1\right\}\)
\(B=\left(\frac{x\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}+\frac{4x+1}{\left(2x-1\right)\left(2x+1\right)}\right).\left(\frac{2x-1}{\left(x+1\right)\left(x^2-x+1\right)}\right)\)
\(=\frac{\left(2x^2+3x+1\right)}{\left(2x+1\right)\left(2x-1\right)}.\frac{\left(2x-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{\left(x+1\right)\left(2x+1\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)\left(x+1\right)\left(x^2-x+1\right)}=\frac{1}{x^2-x+1}\)
a) A có nghĩa <=> \(\left\{{}\begin{matrix}2x-2\ne0\\2-2x^2\ne0\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}x-1\ne0\\\left(1-x\right)\left(x+1\right)\ne0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x\ne1\\x\ne\pm1\end{matrix}\right.\)
b) Ta có:
A = \(\frac{x}{2x-2}+\frac{x^2+1}{2-2x^2}\)
A = \(\frac{x}{2\left(x-1\right)}-\frac{x^2+1}{2\left(x^2-1\right)}\)
A = \(\frac{x\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\frac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
A = \(\frac{x^2+x-x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
A = \(\frac{x+1}{2\left(x-1\right)\left(x+1\right)}=\frac{1}{2\left(x-1\right)}\)
c) A = -1/2
<=> \(\frac{1}{2\left(x-1\right)}=-\frac{1}{2}\)
<=> 2(x - 1) = -2
<=> x - 1 = -1
<=> x = 0 (tmđk)
Vậy x = 0
Mình nghĩ bạn viết hơi sai đề bài.
\(x^2+xz-y^2-yz=\left(x^2-y^2\right)+xz-yz=\left(x-y\right)\left(x+y\right)+z\left(x-y\right)=\left(x-y\right)\left(x+y+z\right)\)
Tương tự: \(y^2+xy-z^2-xz=\left(y-z\right)\left(x+y+z\right)\)
\(z^2+yz-x^2-xy=\left(x+y+z\right)\left(z-x\right)\)
Khi đó:
\(P=\frac{1}{\left(y-z\right)\left(x-y\right)\left(x+y+z\right)}+\frac{1}{\left(z-x\right)\left(y-z\right)\left(x+y+z\right)}+\frac{1}{\left(x-y\right)\left(x+y+z\right)\left(z-x\right)}\)
\(=\frac{z-x+x-y+y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)\left(x+y+z\right)}=0\)
\(5,\)\(\frac{1}{5}x\left(10x-15\right)-2x\left(x-5\right)+7x\)
\(=2x^2-3x+-2x^2+10x-7x\)
\(=0\)
\(\Rightarrow\)Giá trị biểu thức không phụ thuộc vào biến x
\(6,\)\(F=5\left(x^2-3x\right)-x\left(3-5x\right)+18x+3\)
\(=5x^2-15x-3x-5x^2+18x+3\)
\(=3\)
Vậy giá trị biểu thức không phụ thuộc vào biến x
( À có một số chỗ mình phải sửa đề mới đúng đó. Cậu coi lại giùm mình nha )
câu a) mẫu chung: \(x^3x^2\left(x-1\right)^2\) hơi dài nên mk ko làm ^_^
b)
\(=\frac{xy\left(a-b\right)+\left(x-a\right)\left(y-a\right)b-\left(x-b\right)\left(y-b\right)a}{ab\left(a-b\right)}\)
\(=\frac{xya-xyb+bxy-abx-aby+a^2b-axy+axb+aby-ab^2}{ab\left(a-b\right)}\)
\(=\frac{ab\left(a-b\right)}{ab\left(a-b\right)}=1\)
ĐKXĐ: \(\left\{{}\begin{matrix}a-1\ne0\\a^2-1\ne0\\a-a^3\ne0\\a+a^3\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a\ne1\\a\ne\left\{-1;1\right\}\\a\left(1-a^2\right)\ne0\\a\left(1+a^2\right)\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a\ne1\\a\ne\left\{1;-1\right\}\\a\ne\left\{-1;0;1\right\}\\a\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a\ne0\\a\ne-1\\a\ne1\end{matrix}\right.\)
\(M=\frac{a^2}{a-1}+\left(\frac{a}{a^2-1}+\frac{1}{a-a^3}\right):\frac{1-a}{a+a^3}\)
\(=\frac{a^2}{a-1}+\left(\frac{a}{\left(a-1\right)\left(a+1\right)}+\frac{1}{a\left(1-a^2\right)}\right):\frac{1-a}{a\left(1+a^2\right)}\)
\(=\frac{a^2}{a-1}+\left(\frac{a^2}{a\left(a-1\right)\left(a+1\right)}-\frac{1}{a\left(a+1\right)\left(a-1\right)}\right):\frac{1-a}{a\left(1+a^2\right)}\)
\(=\frac{a^2}{a-1}+\frac{\left(a-1\right)\left(a+1\right)}{a\left(a-1\right)\left(a+1\right)}.\frac{a\left(1+a^2\right)}{1-a}\)
\(=\frac{a^2}{a-1}-\frac{1+a^2}{a-1}=\frac{a^2-1-a^2}{a-1}=-\frac{1}{a-1}\)
b/ Thay $a=\frac{1}{2}$ vào M ta được \(M=-\frac{1}{-\frac{1}{2}-1}=-\frac{1}{-\frac{3}{2}}=\frac{1}{\frac{3}{2}}=\frac{2}{3}\)
\(\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}=3\Leftrightarrow x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
Vì \(x,y,z\) khác nhau nên \(x+y+z=0\)
\(P=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\)
\(P=\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}=\frac{\left(-x\right)\cdot\left(-y\right)\cdot\left(-z\right)}{xyz}=-1\)
Vậy...
a/ Đkxđ: \(\left\{{}\begin{matrix}x\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)
Vậy phân thức được xác định khi \(\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)
b/ \(A=\left[1+\frac{1}{x}+\frac{2}{x+1}\left(1+\frac{1}{x}\right)\right]:\frac{x^3+27}{2x}\)
\(=\left[1+\frac{1}{x}+\frac{2}{x+1}+\frac{2}{\left(x+1\right)x}\right]:\frac{\left(x+3\right)\left(x^2-3x+9\right)}{2x}\)
\(=\left[\frac{x\left(x+1\right)+\left(x+1\right)+2x+2}{\left(x+1\right)x}\right].\frac{2x}{\left(x+3\right)\left(x^2-3x+9\right)}\)
\(=\frac{x^2+4x+3}{\left(x+1\right)x}.\frac{2x}{\left(x+3\right)\left(x^2-3x+9\right)}=\frac{\left(x+1\right)\left(x+3\right)}{\left(x+1\right)x}.\frac{2x}{\left(x+3\right)\left(x^2-3x+9\right)}\)
\(=\frac{2}{x^2-3x+9}\)
thk bn nhe