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Bài 2:
a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x-2}\right):\left(\dfrac{x^2-4+16-x^2}{x+2}\right)\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\dfrac{x-x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{12}=\dfrac{-1}{6\left(x-2\right)}\)
b: Thay x=1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(\dfrac{1}{2}-2\right)}=\dfrac{-1}{6\cdot\dfrac{-3}{2}}=\dfrac{1}{9}\)
Thay x=-1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(-\dfrac{1}{2}-2\right)}=-\dfrac{1}{15}\)
c: Để B=2 thì \(\dfrac{-1}{6\left(x-2\right)}=2\)
=>6(x-2)=-1/2
=>x-2=-1/12
hay x=23/12
\(\frac{1}{2}x^2\left(6x-3\right)-x\left(x^2+\frac{1}{2}\right)+\frac{1}{2}\left(x+4\right)\)
\(=\frac{1}{2}x^2.6x+\frac{1}{2}x^2.\left(-3\right)+\left(-x\right).x^2+\left(-x\right).\frac{1}{2}+\frac{1}{2}.x+\frac{1}{2}.4\)
\(=3x^3-\frac{3}{2}x^2-x^3-\frac{1}{2}x+\frac{1}{2}x+2\)
\(=\left(3x^3-x^3\right)-\frac{3}{2}x^2+\left(-\frac{1}{2}x+\frac{1}{2}x\right)+2\)
\(=2x^3-\frac{3}{2}x^2+2\)
\(a,\)\(\frac{1}{2}x^2\left(6x-3\right)-x\left(x^2+\frac{1}{2}\right)+\frac{1}{2}\left(x+4\right)\)
\(=3x^3-\frac{3}{2}x^2-x^3-\frac{1}{2}x+\frac{1}{2}x+2\)
\(=2x^3-\frac{3}{2}x^2+2\)
\(b,\)\(2x\left(3x^3-x\right)-4x^2\left(x-x^2+1\right)+\left(x-3x^2\right)x\)
\(=6x^4-2x^2-4x^3+4x^4-4x^2+x^2-3x^3\)
\(=10x^4-7x^3-5x^2\)
a) ta có :x2+2x+2=(x+1)2+1>0,với mọi x
x2+2x+3=(x+1)2+2>0,với mọi x
ĐKXĐ:x\(\in\)R.Đặt x2+2x+2=a (a>0),ta có:\(\dfrac{a-1}{a}+\dfrac{a}{a+1}=\dfrac{7}{6}\)
<=>\(\dfrac{6\left(a-1\right)\left(a+1\right)}{6a\left(a+1\right)}+\dfrac{6a^2}{6a\left(a+1\right)}=\dfrac{7a\left(a+1\right)}{6a\left(a+1\right)}\)
=>6(a2-1)+6a2=7a2+7a<=>6a2-6+6a2=7a2+7a<=>12a2-7a2-7a-6=0
<=>5a2-7a-6=0<=>(a-2)(5a+3)=0<=>a-2=0(vì a>0,nên 5a+3>0)
<=>a=2=>x2+2x+2=2<=>x(x+2)=0<=>\(|^{x=0}_{x+2=0< =>x=-2}\)
Vậy tặp nghiệm của PT là S\(=\left\{0;-2\right\}\)
Lời giải:
a) ĐKXĐ: $x\neq 0; x\neq \pm 2$
\(A=\left(\frac{x^2}{x(x^2-4)}-\frac{6}{3(x-2)}+\frac{1}{x+2}\right):\frac{(x-2)(x+2)+10-x^2}{x+2}\)
\(=\left(\frac{x}{(x-2)(x+2)}-\frac{2}{x-2}+\frac{1}{x+2}\right):\frac{x^2-4+10-x^2}{x+2}\)
\(=\frac{x-2(x+2)+(x-2)}{(x-2)(x+2)}.\frac{x+2}{6}=\frac{-6}{(x-2)(x+2)}.\frac{x+2}{6}=\frac{1}{2-x}\)
b)
Khi \(|x|=\frac{1}{2}\Rightarrow x=\pm \frac{1}{2}\) (thỏa mãn ĐKXĐ)
\(x=\frac{1}{2}\Rightarrow A=\frac{1}{2-\frac{1}{2}}=\frac{2}{3}\)
\(x=-\frac{1}{2}\Rightarrow A=\frac{1}{2--\frac{1}{2}}=\frac{2}{5}\)
Lời giải:
a) ĐKXĐ: $x\neq 0; x\neq \pm 2$
\(A=\left(\frac{x^2}{x(x^2-4)}-\frac{6}{3(x-2)}+\frac{1}{x+2}\right):\frac{(x-2)(x+2)+10-x^2}{x+2}\)
\(=\left(\frac{x}{(x-2)(x+2)}-\frac{2}{x-2}+\frac{1}{x+2}\right):\frac{x^2-4+10-x^2}{x+2}\)
\(=\frac{x-2(x+2)+(x-2)}{(x-2)(x+2)}.\frac{x+2}{6}=\frac{-6}{(x-2)(x+2)}.\frac{x+2}{6}=\frac{1}{2-x}\)
b)
Khi \(|x|=\frac{1}{2}\Rightarrow x=\pm \frac{1}{2}\) (thỏa mãn ĐKXĐ)
\(x=\frac{1}{2}\Rightarrow A=\frac{1}{2-\frac{1}{2}}=\frac{2}{3}\)
\(x=-\frac{1}{2}\Rightarrow A=\frac{1}{2--\frac{1}{2}}=\frac{2}{5}\)
a) \(\frac{7x}{8}-5\left(x-9\right)=\frac{20x+1,5}{6}\)
\(\Leftrightarrow\frac{7x}{8}-\frac{40\left(x-9\right)}{8}=\frac{20x+1,5}{6}\)
\(\Leftrightarrow\frac{7x}{8}-\frac{40x-360}{8}=\frac{20x+1,5}{6}\)
\(\Leftrightarrow\frac{360-33x}{8}=\frac{20x+1,5}{6}\)
\(\Leftrightarrow2160-198x=160x+12\)
\(\Leftrightarrow358x=2148\)
\(\Leftrightarrow x=6\)
Vậy nghiệm của pt x=6
b) \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{7}-5\)
\(\Leftrightarrow\frac{10\left(x-1\right)+4}{12}-\frac{21x-3}{12}=\frac{4x+2}{7}-\frac{35}{7}\)
\(\Leftrightarrow\frac{-11x-3}{12}=\frac{4x-33}{7}\)
\(\Leftrightarrow-77x-21=48x-396\)
\(\Leftrightarrow125x=375\)
\(\Leftrightarrow3\)
Vậy nghiệm của pt x=3