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Bài 3:
a: \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x-4\right)\left(x+4\right)=21\)
\(\Leftrightarrow x^3-27-x\left(x^2-16\right)=21\)
\(\Leftrightarrow x^3-27-x^3+16x=21\)
=>16x=48
hay x=3
b: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=4\)
\(\Leftrightarrow x^3+8-x^3-2x=4\)
=>-2x=4-8=-4
hay x=2
\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
a: \(=\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}:\left(\dfrac{1}{x+1}+\dfrac{x}{x-1}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}:\dfrac{x-1+x^2+x+2}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{x^2+2x+1}=\dfrac{4x}{x^2+2x+1}\)
b: \(=\dfrac{x+2}{-\left(x-2\right)}\cdot\dfrac{\left(x-2\right)^2}{4x^2}\cdot\left(\dfrac{2}{2-x}-\dfrac{4}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\dfrac{x^2-2x+4}{2-x}\right)\)
\(=\dfrac{-\left(x+2\right)\left(x-2\right)}{4x^2}\cdot\left(\dfrac{2}{2-x}-\dfrac{4}{\left(x+2\right)\left(2-x\right)}\right)\)
\(=\dfrac{-\left(x+2\right)\left(x-2\right)}{4x^2}\cdot\dfrac{2x+4-4}{\left(2-x\right)\left(x+2\right)}\)
\(=\dfrac{2x}{4x^2}=\dfrac{1}{2x}\)
a)
A = \(\left(\dfrac{3-x}{x+3}.\dfrac{x^2+6x+9}{x^2-9}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
=\(\left(\dfrac{3-x}{x+3}.\dfrac{\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}+\dfrac{x}{x+3}\right):\dfrac{3x^3}{x+3}\) (đkxđ: x \(\ne\)\(\pm\)3)
= \(\left(\dfrac{x}{x+3}-1\right).\dfrac{x+3}{3x^2}\)
= \(\dfrac{x-x-3}{x+3}.\dfrac{x+3}{3x^2}\)
= -x2
b) Thay x = \(\dfrac{1}{2}\) vào A, ta có:
A = -\(\left(\dfrac{1}{2}\right)^2\)
= -\(\dfrac{1}{4}\)
c) Để A < 0 thì -x2 < 0
mà -x2 \(\le\) 0 \(\forall\)x
\(\Rightarrow\) Với mọi x (x\(\ne\)0) thì A < 0
a: \(=\dfrac{5}{2}x-2x+\dfrac{7}{2}=\dfrac{1}{2}x+\dfrac{7}{2}\)
b: \(=\dfrac{-1}{4}x^4-3x^2+\dfrac{9}{4}x\)
c: \(=\dfrac{1}{5}x+\dfrac{1}{15}xy+\dfrac{7}{10}x^2\)
d: \(=-9x^3-1-12y+27xy\)
a: x-4=1
=>x=5
Thay x=5 vào B, ta được:
\(B=\dfrac{5+1}{5-3}=\dfrac{6}{2}=3\)
b: \(A=\dfrac{x}{x-3}-\dfrac{x+1}{x+3}+\dfrac{3x-3}{x-3}\)
\(=\dfrac{x+3x-3}{x-3}-\dfrac{x+1}{x+3}\)
\(=\dfrac{4x-3}{x-3}-\dfrac{x+1}{x+3}\)
\(=\dfrac{\left(4x-3\right)\left(x+3\right)-\left(x-3\right)\left(x+1\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{4x^2+12x-3x-9-\left(x^2-2x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{4x^2+9x-9-x^2+2x+3}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x^2+11x-6}{\left(x-3\right)\left(x+3\right)}\)
c: \(M=B:A=\dfrac{3x^2+11x-6}{\left(x-3\right)\left(x+3\right)}:\dfrac{x}{x-3}\)
\(=\dfrac{3x^2+11x-6}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x-3}{x}=\dfrac{3x^2+11x-6}{x^2+3x}\)
M=5
=>\(5\left(x^2+3x\right)=3x^2+11x-6\)
=>\(5x^2+15x-3x^2-11x+6=0\)
=>\(2x^2-4x+6=0\)
=>\(x^2-2x+3=0\)
=>\(\left(x-1\right)^2+2=0\)(vô lý)