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b: Ta có: \(x\left(x+1\right)-\left(2x+3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2x-3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
d: Ta có: \(\left(x-1\right)^2-4\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x-1-2x-4\right)\left(x-1+2x+4\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(3x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-1\end{matrix}\right.\)
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1) \(x^3-8x+7=\left(x-1\right)\left(x^2+x-7\right)\)
2) \(x^3+8x^2-9=\left(x-1\right)\left(x^2+9x+9\right)\)
3) \(3x^3-4x+1=\left(x-1\right)\left(3x^2+3x-1\right)\)
4) \(x^4-3x^2+3x-1=\left(x-1\right)\left(x^3+x^2-2x+1\right)\)
5) \(x^4-5x^2+4=\left(x-1\right)\left(x-2\right)\left(x+1\right)\left(x+2\right)\)
1: Ta có: \(x^3-8x+7\)
\(=x^3-x-7x+7\)
\(=x\left(x-1\right)\left(x+1\right)-7\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x-7\right)\)
2: Ta có: \(x^3+8x^2-9\)
\(=x^3-x^2+9x^2-9\)
\(=x^2\left(x-1\right)+9\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^2+9x+9\right)\)
3: Ta có: \(3x^3-4x+1\)
\(=3x^3-3x-x+1\)
\(=3x\left(x-1\right)\left(x+1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(3x^2+3x-1\right)\)
4: Ta có: \(x^4-3x^2+3x-1\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)-3x\cdot\left(x-1\right)\)
\(=\left(x-1\right)\cdot\left(x^3+x+x^2+1-3x\right)\)
\(=\left(x-1\right)\left(x^3+x^2-2x+1\right)\)
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câu a, \(\dfrac{x}{x+1}\); \(\dfrac{x^2}{1-x}\); \(\dfrac{1}{x^2-1}\) (đk \(x\)≠ -1; 1)
\(x^2\) - 1 = ( \(x\) - 1).(\(x\) + 1)
\(\dfrac{x}{x+1}\) = \(\dfrac{x.\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}\);
\(\dfrac{x^2}{1-x}\) = \(\dfrac{-x^2}{x-1}\)= \(\dfrac{-x^2.\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(\dfrac{1}{x^2-1}\) = \(\dfrac{1}{\left(x-1\right)\left(x+1\right)}\)
b, \(\dfrac{10}{x+2}\); \(\dfrac{5}{2x-4}\); \(\dfrac{1}{6-3x}\) (đk \(x\) ≠ -2; 2)
2\(x-4\) = 2.(\(x\) - 2); 6 - 3\(x\) = - 3.(\(x\) - 2)
\(\dfrac{10}{x+2}\) = \(\dfrac{10.2.3\left(x-2\right)}{2.3\left(x+2\right)\left(x-2\right)}\) = \(\dfrac{60\left(x-2\right)}{6\left(x-2\right)\left(x+2\right)}\)
\(\dfrac{5}{2x-4}\) = \(\dfrac{5.3\left(x+2\right)}{2.3\left(x-2\right).\left(x+2\right)}\) = \(\dfrac{15.\left(x+2\right)}{6.\left(x-2\right)\left(x+2\right)}\)
\(\dfrac{1}{6-3x}\) = \(\dfrac{-1}{3.\left(x-2\right)}\) = \(\dfrac{-1.\left(x+2\right)}{3.2.\left(x-2\right)\left(x+2\right)}\) = \(\dfrac{-2.\left(x+2\right)}{6.\left(x-2\right).\left(x+2\right)}\)
c, \(\dfrac{x}{2x-4}\); \(\dfrac{1}{2x+4}\) và \(\dfrac{3}{4-x^2}\) đk \(x\) ≠ 2; -2
\(\dfrac{x}{2x-4}\) = \(\dfrac{x}{2.\left(x-2\right)}\) = \(\dfrac{x.\left(x+2\right)}{2.\left(x-2\right).\left(x+2\right)}\)
\(\dfrac{1}{2x+4}\) = \(\dfrac{1}{2.\left(x+2\right)}\) = \(\dfrac{\left(x-2\right)}{2.\left(x+2\right).\left(x-2\right)}\)
\(\dfrac{3}{4-x^2}\) = \(\dfrac{-3}{\left(x-2\right)\left(x+2\right)}\) = \(\dfrac{-6}{2.\left(x-2\right)\left(x+2\right)}\)
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a: Xét ΔABC có MN//BC
nên AM/AB=AN/AC
=>AN/4=1,2/3=4/10
hay AN=1,6(cm)
b: BC=5cm
Xét ΔABC có AD là phân giác
nên BD/AB=CD/AC
=>BD/3=CD/4
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{BD}{3}=\dfrac{CD}{4}=\dfrac{BD+CD}{3+4}=\dfrac{5}{7}\)
Do đó: BD=15/7(cm); CD=20/7(cm)
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\(\Leftrightarrow4x^2-12x-4x^2+9=-3\)
=>-12x=-12
hay x=1
\(4x\left(x-3\right)-\left(2x+3\right)\left(2x-3\right)=-3\)
\(4x^2-12x-4x^2+9+3=0\)
\(12-12x=0\\ \Rightarrow1-x=0\\ \Rightarrow x=1\)
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11)11) 3x(x-5)2-(x+2)3+2(x-1)3-(2x+1)(4x2-2x+1)=3x(x2-10x+25)-(x3+6x2+12x+8)+2(x3-3x2+3x-1)-(8x3+1)=3x3-30x2+75x-x3-6x2-12x-8+2x3-6x2+6x-2-8x3-1=-4x3-42x2+63x-11
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Ta có: \(\left(x^{3n}+y^{3n}\right)\left(x^{3n}-y^{3n}\right)=-x^{6n}-y^{6n}\)
\(\Leftrightarrow x^{6n}-y^{6n}=-x^{6n}-y^{6n}\)
\(\Leftrightarrow n\in\varnothing\)
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https://doc-04-90-docs.googleusercontent.com/docs/securesc/4ius91d4i6l7vmjv03uus8n8e1jouf18/nncflvovvggj7929t6q17u30r1267jdr/1638346725000/16593582377474649600/01291187093379538302Z/1HxTe2o4vk4nmppw3mBRNG3Mborq9nJWy?e=download&nonce=fkdac3vjm3rao&user=01291187093379538302Z&hash=p5ovh627ujk6q6qfgi2i0tqtv2b7jgub
Mik chỉ biết dc thế này thui
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Xét ΔMQP có
H,I lần lượt là trung điểm của MQ,MP
=>HI là đường trung bình của ΔMQP
=>HI//QP và HI=QP/2
Xét ΔPMN có
I,K lần lượt là trung điểm của PM,PN
=>IK là đường trung bình của ΔPMN
=>IK//MN và \(IK=\dfrac{MN}{2}\)
b: H,I,K thẳng hàng
mà HI//PQ và IK//MN
nên HI//MN
Ta có: HI//MN
HI//PQ
Do đó: MN//PQ
a: \(x^2-4x-5=\left(x-5\right)\left(x+1\right)\)
b: \(x^2-3x+2=\left(x-2\right)\left(x-1\right)\)
d: \(2x^2-3x+1=\left(x-1\right)\left(2x-1\right)\)
k: \(4x^2-9=\left(2x-3\right)\left(2x+3\right)\)