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30 tháng 11 2017
\(4x^2-25+\left(2x+7\right)\left(5-2x\right)\)
\(=\left(2x-5\right)\left(2x+5\right)+\left(2x+7\right)\left(5-2x\right)\)
\(=\left(2x-5\right)\left(2x+5\right)-\left(2x-7\right)\left(2x-5\right)\)
\(=\left(2x-5\right)\left(2x+5-2x+7\right)\)
\(=\left(2x-5\right).12\)
Những câu khác làm tương tự
TM
30 tháng 5 2017
1) ĐK: \(x\ge-1\)
TH1: \(x^2-3x+1=-x-1\)
\(\Leftrightarrow x^2-2x+2=0\Leftrightarrow\left(x-1\right)^2+1=0\) vô lý
TH2: \(x^2-3x+1=x+1\)
\(\Leftrightarrow x^2-4x=0\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
Vậy ...
N
30 tháng 5 2017
1) \(\left|x^2-3x+1\right|=x+1\)(1)
khi \(x\ge-1\), phương trình (1) có dạng:
\(\orbr{\begin{cases}x^2-3x+1=x+1\\x^2-3x+1=-x-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x^2-4x=0\\x^2-2x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x\left(x-4\right)=0\\\left(x-1\right)^2+1=0\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}\orbr{\begin{cases}x=0\\x=4\end{cases}}\\\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)(vì \(\left(x-1\right)^2+1>0\)(vô nghiệm) )
vậy tập nghiệm của phương trình là: S={0;4}
NN
15 tháng 11 2017
2)
a) \(3x^3-3x=0\)
\(\Leftrightarrow3x\left(x^2-1\right)=0\)
\(\Leftrightarrow3x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Vậy x=0 ; x=-1 ; x=1
b) \(x^2-x+\dfrac{1}{4}=0\)
\(\Leftrightarrow x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{1}{2}=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)
NN
15 tháng 11 2017
1)
a) \(\left(x-2\right)\left(x^2+3x+4\right)\)
\(\Leftrightarrow x^3+3x^2+4x-2x^2-6x-8\)
\(\Leftrightarrow x^3+x^2-2x-8\)
b) \(\left(x-2\right)\left(x-x^2+4\right)\)
\(=x^2-x^3+4x-2x+2x^2-8\)
\(=3x^2-x^3+2x-8\)
c) \(\left(x^2-1\right)\left(x^2+2x\right)\)
\(=x^4+2x^3-x^2-2x\)
d) \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)
\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)
\(=18x^2+12x-9x-6-6x^3-4x^2+3x^2+2x\)
\(=17x^2+5x-6-6x^3\)
15 tháng 12 2020
Tương tự mấy phần kia
\(A=\frac{x+3}{x-2}+\frac{x+2}{3-x}+\frac{x+2}{x^2-5x+6}\)
\(=\frac{x+3}{x-2}-\frac{x+2}{x-3}+\frac{x+2}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}-\frac{\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}+\frac{x+2}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{x^2-9-x^2+4+x+2}{\left(x-2\right)\left(x-3\right)}=\frac{-3+x}{\left(x-2\right)\left(x-3\right)}=\frac{-1}{x-2}\)
NV
8 tháng 1 2018
Bài 2: a) \(3x^3-3x=0\Leftrightarrow3x\left(x^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
b) \(x^2-x+\frac{1}{4}=0\Leftrightarrow x^2-2.\frac{1}{2}+\left(\frac{1}{2}\right)^2=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
24 tháng 2 2017
a, \(\frac{x+16}{49}+\frac{x+18}{47}=\frac{x+20}{45}-1\)
\(\Leftrightarrow1+\frac{x+16}{49}+1+\frac{x+18}{47}=\frac{x+20}{45}-1+2\)
\(\Leftrightarrow\frac{x+16+49}{49}+\frac{x+18+47}{47}=\frac{x+20+45}{45}\)
\(\Leftrightarrow\frac{x+65}{49}+\frac{x+65}{47}-\frac{x+65}{45}=0\)
\(\Leftrightarrow\left(x+65\right)\left(\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\right)=0\)
Ta có: \(\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\)>0
\(\Rightarrow x+65=0\)
\(\Leftrightarrow x=-65\)
Vậy x = -65
b, \(\frac{x-69}{30}+\frac{x-67}{32}+\frac{x-65}{34}=\frac{x-63}{36}+\frac{x-61}{38}+\frac{x-59}{40}\)
\(\Leftrightarrow\frac{x-69}{30}-1+\frac{x-67}{32}-1+\frac{x-65}{34}-1+\frac{x-63}{36}-1+\frac{x-61}{38}-1+\frac{x-59}{40}-1\)
\(\Leftrightarrow\frac{x-99}{30}+\frac{x-99}{32}+\frac{x-99}{34}-\frac{x-99}{36}-\frac{x-99}{38}-\frac{x-99}{40}=0\)
\(\Leftrightarrow\left(x-99\right)\left(\frac{1}{30}+\frac{1}{32}+\frac{1}{34}-\frac{1}{36}-\frac{1}{38}-\frac{1}{40}\right)=0\)
Vì \(\frac{1}{30}+\frac{1}{32}+\frac{1}{34}-\frac{1}{36}-\frac{1}{38}-\frac{1}{40}\)>0
\(\Rightarrow x-99=0\)
\(\Leftrightarrow x=99\)
Vậy x =99
20 tháng 5 2018
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
\(\left(x-2\right)\left(x^2-3x+5\right)=\left(2-x\right)\left(1-x^2\right)\\ \Leftrightarrow\left(x-2\right)\left(x^2-3x+5\right)=-\left(x-2\right)\left(1-x^2\right)\\ \Leftrightarrow\left(x-2\right)\left(x^2-3x+5\right)+\left(x-2\right)\left(1-x^2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x^2-3x+5+1-x^2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(-3x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\-3x+6=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=2\end{matrix}\right.\\ \Leftrightarrow x=2\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-3x+5\right)=-\left(x-2\right)\left(1-x^2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-3x+5\right)+\left(x-2\right)\left(1-x^2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-3x+5+1-x^2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(6-3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=2\end{matrix}\right.\)
Vậy \(S=\left\{2\right\}\)