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a)3x2+4x-9x-12=0
=>(3x2+4x)-(9x+12)=0
=> x(3x+4)-3(3x+4)=0
=> (x-3)(3x+4)=0 =>x-3=0 hoặc 3x+4=0
=>tự tính
b)7x2-9x+2=0
=>7x2-7x-2x+2=0
=>(7x2-7x)-(2x-2)=0
=>7x(x-1)-2(x-1)=0
=>(7x-2)(x-1)=0
=>như câu a
bạn chỉ biết làm 2 câu thôi
a) \(3x^2-5x-12=0\)
\(\Leftrightarrow3x^2+4x-9x-12=0\)
\(\Leftrightarrow x\left(3x+4\right)-3\left(3x+4\right)=0\)
\(\Leftrightarrow\left(3x+4\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x+4=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{4}{3}\\x=3\end{cases}}\)
b) \(7x^2-9x+2=0\)
\(\Leftrightarrow7x^2-7x-2x+2=0\)
\(\Leftrightarrow7x\left(x-1\right)-2\left(x-1\right)=0\).
\(\Leftrightarrow\left(7x-2\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7x-2=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{2}{7}\\x=1\end{cases}}\)
b.
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(a=x^2+7x+11\), ta có:
\(=\left(a-1\right)\left(a+1\right)-24\)
\(=a^2-1-24\)
\(=a^2-25\)
\(=\left(a-5\right)\left(a+5\right)\)
Thay \(a=\left(x^2+7x+11\right)\) ta được:
\(\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)
\(=\left(x^2+7x-6\right)\left(x^2+7x+16\right)\)
a) \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)
\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)
\(\Leftrightarrow-7x+12x=20+2\)
\(\Leftrightarrow5x=22\)
\(\Rightarrow x=\dfrac{22}{5}\)
tick cho mk nha
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)
\(x_1=3;x_2=\dfrac{-11}{10}\)
Tick cho mk nha
Bài 1:
a: \(A=\dfrac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}=\dfrac{x^3\left(x+1\right)+\left(x+1\right)}{x^4-x^3+x^2+x^2-x+1}\)
\(=\dfrac{\left(x+1\right)\left(x^3+1\right)}{\left(x^2-x+1\right)\left(x^2+1\right)}=\dfrac{\left(x+1\right)^2}{x^2+1}\)
Để A=0 thì x+1=0
hay x=-1
b: \(B=\dfrac{x^4-5x^2+4}{x^4-10x^2+9}=\dfrac{\left(x^2-1\right)\left(x^2-4\right)}{\left(x^2-1\right)\left(x^2-9\right)}=\dfrac{x^2-4}{x^2-9}\)
Để B=0 thi (x-2)(x+2)=0
=>x=2 hoặc x=-2
a) \(A=\dfrac{\left(-2\right)^5}{\left(-2\right)^3}=\left(-2\right)^{5-3}=\left(-2\right)^2=4\)
b) \(y\ne0:B=\dfrac{\left(-y\right)^7}{\left(-y\right)^3}=\left(-y\right)^{7-3}=\left(-y\right)^4=y^4\)
c) \(x\ne0:C=\dfrac{\left(x\right)^{12}}{\left(-x\right)^{10}}=\left(x\right)^{12-10}=\left(x\right)^2=x^4\)
d) \(x\ne0:D=\dfrac{2x^6}{\left(2x\right)^3}=\dfrac{2x^6}{8x^3}=\dfrac{1}{4}\left(x\right)^{6-3}=\dfrac{1}{4}\left(x\right)^3\)
e) \(x\ne0:E=\dfrac{\left(-3x\right)^5}{\left(-3x\right)^2}=\left(-3x\right)^{5-2}=\left(-3x\right)^3=-27x^3\)
f) \(x,y\ne0:F=\dfrac{\left(xy^2\right)^4}{\left(xy^2\right)^2}=\left(xy^2\right)^{4-2}=\left(xy^2\right)^2=x^2y^4\)
i) \(x\ne-2:I=\dfrac{\left(x+2\right)^9}{\left(x+2\right)^6}=\left(x+2\right)^{9-6}=\left(x+2\right)^3\)
a: \(3x^2-7x+2\)
\(=3x^2-6x-x+2\)
=(x-2)(3x-1)
b: \(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24\)
\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
c: \(x^4+4=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-4x^2\)
\(=\left(x^2+2x+2\right)\left(x^2-2x+2\right)\)