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\(B=m^2-mn+m-n^2-n+mn=m^2-n^2+n-n\\ =\left(m-n\right)\left(m+n+1\right)\\ =\left(-\dfrac{2}{3}+\dfrac{1}{3}\right)\left(-\dfrac{2}{3}-\dfrac{1}{3}+1\right)=-\dfrac{1}{3}\cdot0=0\)
bài 1:
a) 4n+4+3n-6<19
<=> 7n-2<19
<=> 7n<21 <=> n< 3
b) n\(^2\) - 6n + 9 - n\(^2\) + 16\(\leq\)43
-6n+25\(\leq\)43
-6n\(\leq\)18
n\(\geq\)-3
a: ĐKXĐ: \(x\notin\left\{0;-4;-2;2\right\}\)
b: \(B=\dfrac{1}{x+2}-\dfrac{x^2-4}{x+4}\cdot\left(\dfrac{4x^2+x^2+4x+4}{4x^2\left(x+2\right)^2}\right)\)
\(=\dfrac{1}{x+2}-\dfrac{\left(x-2\right)}{x+4}\cdot\dfrac{5x^2+4x+4}{4x^2\left(x+2\right)}\)
\(=\dfrac{4x^3+16x^2-\left(x-2\right)\left(5x^2+4x+4\right)}{4x^2\left(x+4\right)\left(x+2\right)}\)
\(=\dfrac{4x^3+16x^2-5x^3-4x^2-4x+10x^2+8x+8}{4x^2\left(x+4\right)\left(x+2\right)}\)
\(=\dfrac{-x^3+22x^2+4x+8}{4x^2\left(x+4\right)\left(x+2\right)}\)
B3;a,ĐKXĐ:\(x\ne\pm4\)
A=\(\left(\dfrac{4}{x-4}-\dfrac{4}{x+4}\right)\dfrac{x^2+8x+16}{32}=\left(\dfrac{4x+16}{x^2-16}-\dfrac{4x-16}{x^2-16}\right)\dfrac{x^2+2.4x+4^2}{32}=\left(\dfrac{4x+16-4x+16}{x^2-16}\right)\dfrac{\left(x+4\right)^2}{32}=\left(\dfrac{32}{x^2-16}\right)\dfrac{\left(x+4\right)^2}{32}=\dfrac{32\left(x+4\right)^2}{32.\left(x-4\right)\left(x+4\right)}=\dfrac{x+4}{x-4}\\ \\ \\ \\ \\ \\ b,Tacó\dfrac{x+4}{x-4}=\dfrac{1}{3}\Leftrightarrow3x+12=x-4\Leftrightarrow x=-8\left(TM\right)c,TAcó\dfrac{x+4}{x-4}=3\Leftrightarrow x+4=3x-12\Leftrightarrow x=8\left(TM\right)\)
Bài 1:
a) \(x\ne2\)
Bài 2:
a) \(x\ne0;x\ne5\)
b) \(\dfrac{x^2-10x+25}{x^2-5x}=\dfrac{\left(x-5\right)^2}{x\left(x-5\right)}=\dfrac{x-5}{x}\)
c) Để phân thức có giá trị nguyên thì \(\dfrac{x-5}{x}\) phải có giá trị nguyên.
=> \(x=-5\)
Bài 3:
a) \(\left(\dfrac{x+1}{2x-2}+\dfrac{3}{x^2-1}-\dfrac{x+3}{2x+2}\right)\cdot\left(\dfrac{4x^2-4}{5}\right)\)
\(=\left(\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+3}{2\left(x+1\right)}\right)\cdot\dfrac{2\left(2x^2-2\right)}{5}\)
\(=\dfrac{\left(x+1\right)^2+6-\left(x-1\right)\left(x+3\right)}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{2\cdot2\left(x^2-1\right)}{5}\)
\(=\dfrac{\left(x+1\right)^2+6-\left(x^2+3x-x-3\right)}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{2\left(x-1\right)\left(x+1\right)}{5}\)
\(=\left[\left(x+1\right)^2+6-\left(x^2+2x-3\right)\right]\cdot\dfrac{2}{5}\)
\(=\left[\left(x+1\right)^2+6-x^2-2x+3\right]\cdot\dfrac{2}{5}\)
\(=\left[\left(x+1\right)^2+9-x^2-2x\right]\cdot\dfrac{2}{5}\)
\(=\dfrac{2\left(x+1\right)^2}{5}+\dfrac{18}{5}-\dfrac{2}{5}x^2-\dfrac{4}{5}x\)
\(=\dfrac{2\left(x^2+2x+1\right)}{5}+\dfrac{18}{5}-\dfrac{2}{5}x^2-\dfrac{4}{5}x\)
\(=\dfrac{2x^2+4x+2}{5}+\dfrac{18}{5}-\dfrac{2}{5}x^2-\dfrac{4}{5}x\)
\(=\dfrac{2x^2+4x+2+18}{5}-\dfrac{2}{5}x^2-\dfrac{4}{5}x\)
\(=\dfrac{2x^2+4x+20}{5}-\dfrac{2}{5}x^2-\dfrac{4}{5}x\)
c) tự làm, đkxđ: \(x\ne1;x\ne-1\)
ô hô ngộ quá nhìu người bt toán lớp 8 trong khi lớp 7 với lại óc nguyow trở lại r kaka
a) ĐKXĐ : \(x\ne\pm3\)
b) \(A=\left(\dfrac{1}{x+3}-\dfrac{1}{3-x}\right):\left(2-\dfrac{6}{3-x}\right)\)
\(A=\dfrac{\left(\dfrac{3-x}{\left(x+3\right)\left(3-x\right)}-\dfrac{x+3}{\left(3-x\right)\left(x+3\right)}\right)}{\left(\dfrac{2\left(3-x\right)}{3-x}-\dfrac{6}{3-x}\right)}\)
\(A=\dfrac{\left(\dfrac{3-x-x-3}{\left(x+3\right)\left(3-x\right)}\right)}{\left(\dfrac{6-2x-6}{3-x}\right)}\)
\(A=\dfrac{\left(\dfrac{-2x}{\left(x+3\right)\left(3-x\right)}\right)}{\left(\dfrac{-2x}{3-x}\right)}\)
\(A=\dfrac{-2x}{\left(x+3\right)\left(3-x\right)}.\dfrac{3-x}{-2x}\)
\(A=\dfrac{\left(-2x\right).\left(3-x\right)}{\left(x+3\right)\left(3-x\right).\left(-2x\right)}\)
\(\Leftrightarrow A=\dfrac{1}{x+3}\)
c) Thay \(A=\dfrac{1}{6}\) ta có :
\(\dfrac{1}{x+3}=\dfrac{1}{6}\)
\(x+3=1:\dfrac{1}{6}\)
\(x+3=6\)
x=6-3
x=3
d) \(A=\dfrac{1}{x+3}\)
=> x+3 thuộc Ư(1)={-1,1}
=> x thuộc {-4,-2}
\(B=m\left(m-n+1\right)-n\left(n+1-m\right)=m^2-mn+m-n^2-n+mn=m^2-n^2+m-n=\left(m-n\right)\left(m+n\right)+\left(m-n\right)=\left(m-n\right)\left(m+n+1\right)=\left(-\dfrac{2}{3}+\dfrac{1}{3}\right)\left(-\dfrac{2}{3}-\dfrac{1}{3}+1\right)=-\dfrac{1}{3}.0=0\)