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(3x+2)^2-(x-6)^2=0
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DD
1
7 tháng 11 2016
a, 3x(x^2-4)=0
+3x=0=>x=0
+x^2-4=0
=>x^2=4
=>x=+-2
c,x(x+2)-3x-6=0
x(x+2)-3(x+2)=0
(x+2)(x-3)=0
TH1 :x+2=0
x=-2
TH2 : x-3=0
x=3
câu b bạn chờ mình chúc nha
nhớ k cho mình
2 tháng 9 2021
(3x+2)2-(x-6)2=(3x+2-x+6)(3x+2+x-6)=(2x+8)(4x-4)=8(x+4)(x-1)
AT
1
11 tháng 8 2021
a: Ta có: \(2-x=2\left(x-2\right)^3\)
\(\Leftrightarrow2\left(x-2\right)^3+x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left[2\left(x-2\right)^2+1\right]=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
c: Ta có: \(\left(x-1.5\right)^6+2\left(1.5-x\right)^3=0\)
\(\Leftrightarrow\left(x-1.5\right)^6-2\left(x-1.5\right)^3=0\)
\(\Leftrightarrow\left(x-1.5\right)^3\cdot\left[\left(x-1.5\right)^3-2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1.5\\x=\sqrt[3]{2}+1.5\end{matrix}\right.\)
HN
2
NN
31 tháng 10 2020
Đặt \(x^2+3x+1=t\)
\(\Rightarrow\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6=t.\left(t+1\right)-6\)
\(=t^2+t-6=\left(t^2-2t\right)+\left(3t-6\right)\)
\(=t\left(t-2\right)+3\left(t-2\right)=\left(t-2\right)\left(t+3\right)\)
\(=\left(x^2+3x+1-2\right)\left(x^2+3x+1+3\right)\)
\(=\left(x^2+3x-1\right)\left(x^2+3x+4\right)\)
G
31 tháng 10 2020
\(A=\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)
Đặt \(x^2+3x+1=a\)ta có :
\(a\left(a+1\right)-6\)
\(=a^2+a-6\)
\(=a^2+6a-a-6\)
\(=\left(a^2+6a\right)-\left(a+6\right)\)
\(=a\left(a+6\right)-\left(a+6\right)\)
\(=\left(a+6\right)\left(a-1\right)\)
Thay \(a=x^2+3x+1\)vào A ta có :
\(A=\left(x^2+3x+1+6\right)\left(x^2+3x+1-1\right)\)
\(=\left(x^2+3x+7\right)\left(x^2+3x\right)\)
BA
1
9 tháng 7 2019
\(\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)
Đặt \(\left(x^2+3x+1\right)=a\), ta được:
\(a\left(a+1\right)-6\)\(=a^2+a-6\)\(=\left(a^2+3a\right)-\left(2a+6\right)\)\(=a\left(a+3\right)-2\left(a+3\right)\)
\(=\left(a+3\right)\left(a-2\right)\)
Thay \(a=\left(x^2+3x+1\right)\), ta được:
\(=\left(x^2+3x+1+3\right)\left(x^2+3x+1-2\right)\)
\(=\left(x^2+3x+4\right)\left(x^2+3x-1\right)\)
Y
23 tháng 6 2023
\(\left(5x-10\right)\left(x^2-1\right)-\left(3x-6\right)\left(x^2-2x+1\right)\)
\(=\left(5x-10\right)\left(x-1\right)\left(x+1\right)-\left(3x-6\right)\left(x-1\right)^2\)
\(=\left(x-1\right)\left[\left(5x-10\right)\left(x+1\right)-\left(3x-6\right)\left(x-1\right)\right]\)
\(=\left(x-1\right)\left[5\left(x-2\right)\left(x+1\right)-3\left(x-2\right)\left(x-1\right)\right]\)
\(=\left(x-1\right)\left[\left(x-2\right)\left(5x+5-3x+3\right)\right]\)
\(=\left(x-1\right)\left[\left(x-2\right)\left(2x+8\right)\right]\)
\(=\left(x-1\right)\left(x-2\right)\left(2x+8\right)\)
T
21 tháng 10 2023
\(a,x^2-5x+6\\=x^2-3x-2x+6\\=x(x-3)-2(x-3)\\=(x-3)(x-2)\\---\\b,3x^2+9x-30\\=3x^2-6x+15x-30\\=3x(x-2)+15(x-2)\\=(x-2)(3x+15)\\=3(x-2)(x+5)\\---\)
\(c,x^2-3x+2\\=x^2-x-2x+2\\=x(x-1)-2(x-1)\\=(x-1)(x-2)\\---\\d,3x^2-5x-2\\=3x^2-6x+x-2\\=3x(x-2)+(x-2)\\=(x-2)(3x+1)\\Toru\)
\(\left(3x+2\right)^2-\left(x-6\right)^2=\left(3x+2-x+6\right)\left(3x+2+x-6\right)=\left(2x+8\right)\left(4x-4\right)=8\left(x+4\right)\left(x-1\right)\)
\(\left(3x+2\right)^2-\left(x-6\right)^2\)
\(=\left(3x+2+x-6\right)\left(3x+2-x+6\right)\)
\(=\left(4x-4\right)\left(2x+8\right)\)
\(=8\left(x-1\right)\left(x+4\right)\)