Phân tích thành nhân tử :
3+4x (với x < 0)
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1 tháng 8 2016
\(x-5\)
\(=\left(\sqrt{x}\right)^2-\left(\sqrt{5}\right)^2\)
\(=\left(\sqrt{x}-\sqrt{5}\right)\left(\sqrt{x}+\sqrt{5}\right)\)
21 tháng 4 2021
\(5-7x^2=\left(\sqrt{5}\right)^2-\left(x\sqrt{7}\right)^2\)
\(=\left(\sqrt{5}-x\sqrt{7}\right)\left(\sqrt{5}+x\sqrt{7}\right)\)
\(3+4x=\left(\sqrt{3}\right)^2-\left(2\sqrt{x}\right)^2\) ( do x<0 )
\(=\left(\sqrt{3}-2\sqrt{x}\right)\left(3+2\sqrt{x}\right)\)
DN
1
12 tháng 12 2023
a: \(x^2+12x+36=0\)
=>\(x^2+2\cdot x\cdot6+6^2=0\)
=>\(\left(x+6\right)^2=0\)
=>x+6=0
=>x=-6
b: \(4x^2-4x+1=0\)
=>\(\left(2x\right)^2-2\cdot2x\cdot1+1^2=0\)
=>\(\left(2x-1\right)^2=0\)
=>2x-1=0
=>2x=1
=>x=1/2
c: \(x^3+6x^2+12x+8=0\)
=>\(x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=0\)
=>\(\left(x+2\right)^3=0\)
=>x+2=0
=>x=-2
6 tháng 11 2021
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
VN
2
4 tháng 11 2017
pt <=> (x^3-x^2) - (3x^2-3x) +(2x-2) = 0
<=> (x-1).(x^2-3x+2) = 0
<=>(x-1).[(x^2-x) - (2x-2)] = 0
<=> (x-1)^2 . (x-2) = 0
<=> x-1 = 0 hoặc x-2 = 0
<=> x=1 hoặc x=2
OP
9 tháng 8 2016
\(\left(x+5\right)\left(x-5\right)-\left(x-2\right)\left(x+7\right)=0\)
\(\left(x^2-5^2\right)-\left(x^2+7x-2x-14\right)=0\)
\(x^2-25-x^2-7x+2x+14=0\)
\(-5x=25-14\)
\(-5x=11\)
\(x=-\frac{11}{5}\)
***
\(9x^2-4-2\left(3x-2\right)^2=0\)
\(\left(3x\right)^2-2^2-2\left(3x-2\right)^2=0\)
\(\left(3x-2\right)\left(3x+2\right)-2\left(3x-2\right)^2=0\)
\(\left(3x-2\right)\left[\left(3x+2\right)-2\left(3x-2\right)\right]=0\)
\(\left(3x-2\right)\left(3x+2-6x+4\right)=0\)
\(\left(3x-2\right)\left(6-3x\right)=0\)
TH1:
\(3x-2=0\)
\(3x=2\)
\(x=\frac{2}{3}\)
TH2:
\(6-3x=0\)
\(3x=6\)
\(x=\frac{6}{3}\)
\(x=2\)
Vậy \(x=\frac{2}{3}\) hoặc \(x=2\)
***
\(12\left(3-4x\right)+7\left(4x-3\right)=0\)
\(12\left(3-4x\right)-7\left(3-4x\right)=0\)
\(\left(3-4x\right)\left(12-7\right)=0\)
\(5\left(3-4x\right)=0\)
\(3-4x=0\)
\(4x=3\)
\(x=\frac{3}{4}\)
***
\(x^2-4-2xy+y^2=\left(x-y\right)^2-2^2=\left(x-y-2\right)\left(x-y+2\right)\)
***
\(x^3-4x^2-12x+27=\left(x+3\right)\left(x^2-3x+9\right)-4x\left(x+3\right)=\left(x+3\right)\left(x^2-3x+9-4x\right)=\left(x+3\right)\left(x^2-7x+9\right)\)
***
\(3x^2-18x+27=3\left(x^2-2\times x\times3+3^2\right)=3\left(x-3\right)^2\)
***
\(A=-x^2+3x-4=-\left(x^2-2\times x\times\frac{3}{2}+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+4\right)=-\left[\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\right]\)
\(\left(x-\frac{3}{2}\right)^2\ge0\)
\(\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}\)
\(-\left[\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\right]\le-\frac{7}{4}< 0\)
Vậy A < 0 với mọi x (đpcm)
9 tháng 8 2016
1a (x+5)(x-5)-(x-2)(x+7) = 0
=> x2-25-(x2+5x-14) = 0
=> x2-25-x2-5x+14 = 0
=> -11-5x = 0
=> -5x = -11-0
=> -5x = -11
=> x = -11:5
=> x = \(\frac{-11}{5}\)
bài 2:
1) (x-y)2-4
3) 3(x2-6x+9)