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a) Ta có 2 u 2 − 20 u + 50 5 u + 5 . 2 u 2 − 2 4 ( u − 5 ) 3 = 2 ( u − 5 ) 2 5 ( u + 1 ) . 2 ( u − 1 ) ( u + 1 ) 4 ( u − 5 ) 3 = u − 1 5 ( u − 5 )
b) Ta có v + 3 v 2 − 4 . 8 − 12 v + 6 v 2 − v 3 7 v + 21 = v + 3 ( v − 2 ) ( v + 2 ) . ( 2 − v ) 3 7 ( v + 3 ) = 1 ( v − 2 ) ( v + 2 ) . − ( v − 2 ) 3 7 = − ( v − 2 ) 2 7 ( v + 2 )
Bài 1:
a) Ta có: \(VT=\frac{-u^2+3u-2}{\left(u+2\right)\left(u-1\right)}\)
\(=\frac{-\left(u^2-3u+2\right)}{\left(u+2\right)\left(u-1\right)}\)
\(=\frac{-\left(n^2-u-2u+2\right)}{\left(u+2\right)\left(u-1\right)}\)
\(=\frac{-\left[u\left(u-1\right)-2\left(u-1\right)\right]}{\left(u+2\right)\left(u-1\right)}\)
\(=\frac{-\left(u-1\right)\left(u-2\right)}{\left(u+2\right)\left(u-1\right)}\)
\(=\frac{2-u}{u+2}\)(1)
Ta có: \(VP=\frac{u^2-4u+4}{4-u^2}\)
\(=\frac{\left(u-2\right)^2}{-\left(u-2\right)\left(u+2\right)}\)
\(=\frac{-\left(u-2\right)}{u+2}\)
\(=\frac{2-u}{u+2}\)(2)
Từ (1) và (2) suy ra \(\frac{-u^2+3u-2}{\left(u+2\right)\left(u-1\right)}=\frac{u^2-4u+4}{4-u^2}\)
b) Ta có: \(VT=\frac{v^3+27}{v^2-3v+9}\)
\(=\frac{\left(v+3\right)\left(v^3-3u+9\right)}{v^2-3u+9}\)
\(=v+3=VP\)(đpcm)
Bài 2:
a) Ta có: \(\frac{3x^2-2x-5}{M}=\frac{3x-5}{2x-3}\)
\(\Leftrightarrow\frac{3x^2-5x+3x-5}{M}=\frac{3x-5}{2x-3}\)
\(\Leftrightarrow\frac{x\left(3x-5\right)+\left(3x-5\right)}{M}=\frac{3x-5}{2x-3}\)
\(\Leftrightarrow\frac{\left(3x-5\right)\left(x+1\right)}{M}=\frac{3x-5}{2x-3}\)
\(\Leftrightarrow M=\frac{\left(3x-5\right)\left(x+1\right)\left(2x-3\right)}{3x-5}\)
\(\Leftrightarrow M=\left(x+1\right)\left(2x-3\right)\)
\(\Leftrightarrow M=2x^2-3x+2x-3\)
hay \(M=2x^2-x-3\)
Vậy: \(M=2x^2-x-3\)
b) Ta có: \(\frac{2x^2+3x-2}{x^2-4}=\frac{M}{x^2-4x+4}\)
\(\Leftrightarrow\frac{2x^2+4x-x-2}{\left(x-2\right)\left(x+2\right)}=\frac{M}{\left(x-2\right)^2}\)
\(\Leftrightarrow\frac{2x\left(x+2\right)-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{M}{\left(x-2\right)^2}\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(2x-1\right)}{\left(x+2\right)\left(x-2\right)}=\frac{M}{\left(x-2\right)^2}\)
\(\Leftrightarrow\frac{M}{\left(x-2\right)^2}=\frac{2x-1}{x-2}\)
\(\Leftrightarrow M=\frac{\left(2x-1\right)\left(x-2\right)^2}{\left(x-2\right)}\)
\(\Leftrightarrow M=\left(2x-1\right)\left(x-2\right)\)
\(\Leftrightarrow M=2x^2-4x-x+2\)
hay \(M=2x^2-5x+2\)
Vậy: \(M=2x^2-5x+2\)
Bài 3:
a) Ta có: \(\frac{x+1}{N}=\frac{x^2-2x+4}{x^3+8}\)
\(\Leftrightarrow\frac{x+1}{N}=\frac{x^2-2x+4}{\left(x+2\right)\left(x^2-2x+4\right)}\)
\(\Leftrightarrow\frac{x+1}{N}=\frac{1}{x+2}\)
\(\Leftrightarrow N=\left(x+1\right)\left(x+2\right)\)
hay \(N=x^2+3x+2\)
Vậy: \(N=x^2+3x+2\)
n) Ta có: \(\frac{\left(x-3\right)\cdot N}{3+x}=\frac{2x^3-8x^2-6x+36}{2+x}\)
\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=\frac{2x^3+4x^2-12x^2-24x+18x+36}{x+2}\)
\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{\left(x+3\right)}=\frac{2x^2\left(x+2\right)-12x\left(x+2\right)+18\left(x+2\right)}{x+2}\)
\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=\frac{\left(x+2\right)\left(2x^2-12x+18\right)}{x+2}\)
\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=2x^2-12x+18\)
\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=2x^2-6x-6x+18=2x\left(x-3\right)-6\left(x-3\right)=2\cdot\left(x-3\right)^2\)
\(\Leftrightarrow N\cdot\left(x-3\right)=\frac{2\left(x-3\right)^2}{x+3}\)
\(\Leftrightarrow N=\frac{2\left(x-3\right)^2}{x+3}:\left(x-3\right)=\frac{2\left(x-3\right)^2}{\left(x+3\right)\left(x-3\right)}\)
\(\Leftrightarrow N=\frac{2\left(x-3\right)}{x+3}\)
hay \(N=\frac{2x-6}{x+3}\)
Vậy: \(N=\frac{2x-6}{x+3}\)
a) ∀ x , y ∈ ℝ
b) Chú ý: A 2 + B 2 ≥ 0 với ∀ A , B . Dấu "=" xảy ra khi A = 0 B = 0
Từ đó tìm được điều kiện xác định là: u ≠ 1 và v ≠ -2.
a, \(x^2+2x\left(y+1\right)+y^2+2y+1=\left(x^2+2xy+y^2\right)+\left(2x+2y\right)+1\)
\(=\left(x+y\right)^2+2\left(x+y\right)+1=\left(x+y+1\right)^2\)
b, \(u^2+v^2+2u+2v+2\left(u+1\right)\left(v+1\right)+2\)
\(=u^2+v^2+2u+2v+2uv+2u+2v+2+2\)
\(=\left(u^2+2uv+v^2\right)+\left(4u+4v\right)+4\)
\(=\left(u+v\right)^2+4\left(u+v\right)+2^2=\left(u+v+2\right)^2\)
1.
a) \(A=x^2+2x\left(y+1\right)+y^2+2y+1\)
\(A=x^2+2x\left(y+1\right)+\left(y+1\right)^2\)
\(A=\left(x+y+1\right)^2\)
b) \(B=u^2+v^2+2u+2v+2\left(u+1\right)\left(v+1\right)+2\)\(B=u^2+v^2+2u+2v+2\left(u+1\right)\left(v+1\right)+1+1\)\(B=\left(u^2+2u+1\right)+2\left(u+1\right)\left(v+1\right)+\left(v^2+2v+1\right)\)\(B=\left(u+1\right)^2+2\left(u+1\right)\left(v+1\right)+\left(v+1\right)^2\)\(B=\left(u+1+v+1\right)^2=\left(u+v+2\right)^2\)
tik mik nha !!!
a. A = (a + b)3 - (a - b)3
A = \(\left[\left(a+b\right)-\left(a-b\right)\right]\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
A = (a + b - a + b)\(\left[a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right]\)
A = 2b(a2 + a2 + a2 + 2ab - 2ab + b2 - b2 + b2)
A = 2b(3a2 + b2)
A = 6a2b + 2b3
a) A = u 3 + 6 uv 2 – v 3 .
b) B = ( c + 2 d ) + ( c − 2 d 3 = 8 c 3 .
a) 4x^2(5x^3 - 2x + 3)
= 20x^5 - 8x^3 + 12x^2
b) 2u(1 + u - v) - v(1 - 2u + v)
= 2u + 2u^2 - v - v^2
a) b 3 + 3 b 2 + 2 b 3 + 1 . b) 0.