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a) \(A=\left(3x-2\right)^2+\left(x+1\right)^2-2\left(x+1\right)\left(3x-2\right)\)
\(\Leftrightarrow A=\left(x+1\right)^2-2\left(x+1\right)\left(3x-2\right)+\left(3x-2\right)^2\)
\(\Leftrightarrow A=\left[\left(x+1\right)-\left(3x-2\right)\right]^2\)
\(\Leftrightarrow A=\left(x+1-3x+2\right)^2\)
\(\Leftrightarrow A=\left(3-2x\right)^2\)
Thay \(x=\dfrac{3}{2}\) vào biểu thức A ta được:
\(\left(3-2.\dfrac{3}{2}\right)^2=\left(3-3\right)^2=0^2=0\)
Vậy giá trị của biểu thức A tại \(x=\dfrac{3}{2}\) là 0
b) \(B=\dfrac{x^2y\left(y-x\right)-xy^2\left(x-y\right)}{3y^2-3x^2}\)
\(\Leftrightarrow B=\dfrac{x^2y\left(y-x\right)+xy^2\left(y-x\right)}{3\left(y^2-x^2\right)}\)
\(\Leftrightarrow B=\dfrac{\left(y-x\right)\left(x^2y+xy^2\right)}{3\left(y-x\right)\left(y+x\right)}\)
\(\Leftrightarrow B=\dfrac{xy\left(y-x\right)\left(x+y\right)}{3\left(y-x\right)\left(y+x\right)}\)
\(\Leftrightarrow B=\dfrac{xy\left(y-x\right)\left(y+x\right)}{3\left(y-x\right)\left(y+x\right)}\)
\(\Leftrightarrow B=\dfrac{xy}{3}\)
Thay \(x=-3\) và \(y=\dfrac{1}{2}\) vào biểu thức B ta được:
\(\dfrac{\left(-3\right).\dfrac{1}{2}}{3}=\dfrac{\dfrac{-3}{2}}{3}=\dfrac{\dfrac{-3}{2}}{3}=\dfrac{-1}{2}\)
Vậy giá trị của biểu thức B tại \(x=-3\) và \(y=\dfrac{1}{2}\) là \(\dfrac{-1}{2}\)
c) \(C=\dfrac{x+1}{x-3}-\dfrac{1-x}{x+3}-\dfrac{2x\left(1-x\right)}{9-x^2}\)
\(\Leftrightarrow C=\dfrac{x+1}{x-3}-\dfrac{1-x}{x+3}+\dfrac{2x\left(1-x\right)}{x^2-9}\)
\(\Leftrightarrow C=\dfrac{x+1}{x-3}-\dfrac{1-x}{x+3}+\dfrac{2x\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}\) MTC: \(\left(x-3\right)\left(x+3\right)\)
\(\Leftrightarrow C=\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow C=\dfrac{\left(x+1\right)\left(x+3\right)-\left(x-3\right)\left(1-x\right)+2x\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow C=\dfrac{\left(x^2+3x+x+3\right)-\left(x-x^2-3+3x\right)+\left(2x-2x^2\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow C=\dfrac{x^2+3x+x+3-x+x^2+3-3x+2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow C=\dfrac{2x+6}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow C=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow C=\dfrac{2}{x-3}\)
Thay \(x=5\) vào biểu thức C ta được:
\(\dfrac{2}{5-3}=\dfrac{2}{2}=1\)
Vậy giá trị của biểu thức C tại \(x=5\) là 1
a: \(=\dfrac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{2\left(x-1\right)\left(x+1\right)}{5}\cdot2\)
\(=\dfrac{10}{5}\cdot2=4\)
b: \(=\dfrac{x}{x-3}-\dfrac{x\left(x+3\right)}{2x+3}\cdot\dfrac{x^2+6x+9-x^2}{x\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x}{x-3}-\dfrac{3}{x-3}=1\)
1) Đặt \(B=x^2+y^2+z^2\)
\(C=\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)\)
Ta có: \(x+y+z=0\Rightarrow\left(x+y+z\right)^2=0\)
\(\Leftrightarrow-2\left(xy+yz+xz\right)=x^2+y^2+z^2\)
Suy ra: \(C=2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)=2\left(x^2+y^2+z^2\right)+x^2+y^2+z^2=3\left(x^2+y^2+z^2\right)\)
\(\Rightarrow A=\dfrac{B}{C}=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1}{3}\)
2) \(x^2-2y^2=xy\Leftrightarrow x^2-xy-2y^2=0\)
\(\Leftrightarrow x^2+xy-2xy-2y^2=0\)
\(\Leftrightarrow x\left(x+y\right)-2y\left(x+y\right)=0\)
\(\Leftrightarrow\left(x-2y\right)\left(x+y\right)=0\)
Do \(x+y\ne0\) nên \(x-2y=0\Leftrightarrow x=2y\)
Do đó: \(A=\dfrac{2y-y}{2y+y}=\dfrac{y}{3y}=\dfrac{1}{3}\)
\(a,\dfrac{1}{x^2-x}+\dfrac{2x}{4x^3}-\dfrac{1}{x^2+x+1}\)
\(=\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{2x^2}-\dfrac{1}{x^2+x+1}\)
\(=\dfrac{2x\left(x^2+x+1\right)+\left(x-1\right).\left(x^2+x+1\right)-2x^2.\left(x-1\right)}{2x^2.\left(x-1\right).\left(x^2+x+1\right)}\)
\(=\dfrac{2x^3+2x^2+2x+x^3-1-2x^3+2x^2}{2x^2.\left(x^3-1\right)}\)
\(=\dfrac{4x^2+2x+x^3-1}{2x^5-2x^2}\)
\(=\dfrac{x^3+4x^2+2x-1}{2x^5-2x^2}\)
\(b,\dfrac{1}{x^2-x+1}+1-\dfrac{x^2+2}{\left(x+1\right).\left(x^2-x+1\right)}\)
\(=\dfrac{1}{x^2-x+1}+1-\dfrac{x^2+2}{\left(x^2-x+1\right)}\)
\(=\dfrac{x+1\left(x+1\right).\left(x^2-x+1\right)-\left(x^2+2\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x+1+x^3+1-x^2-2}{\left(x+1\right).\left(x^2-x+1\right)}\)
\(=\dfrac{x+0+x^3-x^2}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x\left(1+x^2-x\right)}{\left(x+1\right).\left(x^2-x+1\right)}\)
\(=\dfrac{x}{x+1}\)
1. Đặt : x = a + \(\dfrac{1}{3}\) ; y = b + \(\dfrac{1}{3}\) ; z = \(c+\dfrac{1}{3}\)
Ta có : x + y + z = 1
⇒ a + b + c = 0
Ta có : x2 + y2 + z2 = ( a + \(\dfrac{1}{3}\))2 + ( b + \(\dfrac{1}{3}\))2 + ( c + \(\dfrac{1}{3}\))2
= a2 + \(\dfrac{2}{3}a+\dfrac{1}{9}+b^2+\dfrac{2}{3}b+\dfrac{1}{9}+c^2+\dfrac{2}{3}c+\dfrac{1}{9}\)
= \(\dfrac{1}{3}+\dfrac{2}{3}\left(a+b+c\right)+a^2+b^2+c^2\)
= \(\dfrac{1}{3}+a^2+b^2+c^2\) ≥ \(\dfrac{1}{3}\)
Dâu "=" xảy ra khi và chỉ khi : a = b = c = 0 ⇔ x = y = z = \(\dfrac{1}{3}\)
b) \(\frac{4}{x+2}+\frac{3}{x-2}+\frac{5x+2}{4-x^2}\left(x\ne\pm2\right)\)
\(=\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{5x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{4x-8+3x+6-5x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2x}{\left(x-2\right)\left(x+2\right)}\)
f) \(x^2+1-\frac{x^4-3x^2+2}{x^2-1}\)
\(=x^2+1-\frac{\left(x^2-2\right)\left(x^2-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=x^2+1-\frac{\left(x^2-2\right)\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=x^2+1-\left(x^2-2\right)\)
\(=x^2+1-x^2+2\)
\(=3\)
1.
c) x2 - xy - 3x + 3y
= (x2 - xy) - (3x - 3y)
= x(x - y) - 3(x - y)
= (x - 3)(x - y)
3.
ĐKXĐ: \(x\ne y,y\ne z,z\ne x\)
Ta có:
\(\dfrac{1}{\left(x-y\right)\left(y-z\right)}+\dfrac{1}{\left(y-z\right)\left(z-x\right)}+\dfrac{1}{\left(z-x\right)\left(x-y\right)}=\dfrac{\left(z-x\right)+\left(x-y\right)+\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=0\)