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a) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x\right)^2-5^2-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(-2\right).\left(2x-5\right)=0\)
\(\Leftrightarrow2x-5=0\)
\(\Leftrightarrow x=\dfrac{5}{2}\)
a,\(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left(4x^2-25\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left(2x-5\right)^2-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(2x-5-2x-7\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(-12\right)=0\)
\(\Rightarrow2x-5=0\)
\(\Rightarrow2x=5\)
\(\Rightarrow x=\dfrac{5}{2}\)
\(b,2x^3+3x^2+2x+3=0\)
\(\Rightarrow\left(2x^3+2x\right)+\left(3x^2+3\right)=0\)
\(\Rightarrow2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)
\(\Rightarrow\left(2x+3\right)\left(x^2+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x+3=0\\x^2+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=-3\\x^2=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=1\end{matrix}\right.\)
\(c,x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x^3+27\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x+3\right)^3+\left(x+3\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2+9+x-9\right)=0\)
\(\Rightarrow\left(x+3\right).x^3=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x^3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=0\end{matrix}\right.\)
\(d,x^2\left(x+7\right)-4\left(x+7\right)=0\)
\(\Rightarrow\left(x^2-4\right)\left(x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-4=0\\x+7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2=4\\x=-7\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\)
Dài dữ trời :V Về sau gửi từng bài một thôi, nhìn hoa mắt quá @@
B1: Phân tích thành nhân tử:
a) \(6x^2+9x=3x\left(2x+3\right)\)
b) \(4x^2+8x=4x\left(x+2\right)\)
c) \(5x^2+10x=5x\left(x+2\right)\)
d) \(2x^2-8x=2x\left(x-4\right)\)
e) \(5x-15y=5\left(x-3y\right)\)
f) \(x\left(x^2-1\right)+3\left(x^2-1\right)=\left(x^2-1\right)\left(x+3\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+3\right)\)
g) \(x^2-2x+1-4y^2=\left(x-1\right)^2-4y^2\)
\(=\left(x-1-2y\right)\left(x-1+2y\right)\)
h) \(x^2-100=\left(x-10\right)\left(x+10\right)\)
i) \(9x^2-18x+9=\left(3x-3\right)^2\)
k) \(x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)
l) \(x^2+6xy^2+9y^4=\left(x+3y\right)^2\)
m) \(4xy-4x^2-y^2=-\left(4x^2-4xy+y^2\right)\)
\(=-\left(2x-y\right)^2\)
n) \(\left(x-15\right)^2-16=\left(x-15-16\right)\left(x-15+16\right)\)
\(=\left(x-31\right)\left(x+1\right)\)
o) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3+x\right)\)
\(=\left(2+x\right)\left(8+x\right)\)
p) \(\left(7x-4\right)^2-\left(2x+1\right)^2\)
\(=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)\)
\(=\left(5x-5\right)\left(9x-3\right)\)
Bài 1 :
a ) \(6x^2+9x=3x\left(x+3\right)\)
b ) \(4x^2+8x=4x\left(x+2\right)\)
c ) \(5x^2+10x=5x\left(x+2\right)\)
d ) \(2x^2-8x=2x\left(x-4\right)\)
e ) \(5x-15y=5\left(x-3y\right)\)
f ) \(x\left(x^2-1\right)+3\left(x^2-1\right)=\left(x^2-1\right)\left(x+3\right)\)
g ) \(x^2-2x+1-4y^2=\left(x-1\right)^2-\left(2y\right)^2=\left(x-1-2y\right)\left(x-1+2y\right)\)
h ) \(x^2-100=x^2-10^2=\left(x-10\right)\left(x+10\right)\)
i ) \(9x^2-18x+9=\left(3x-3\right)^2\)
k ) \(x^3-8=\left(x-2\right)\left(x^2+2x+2^2\right)\)
l ) \(x^2+6xy^2+9y^4=\left(x+3y^2\right)^2\)
m ) \(4xy-4x^2-y^2=-\left(2x-y\right)^2\)
n ) \(\left(x-15\right)^2=x^2-30x+15^2\)
o ) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3-x\right)=\left(2+x\right)\left(8-x\right)\)
p ) \(\left(7x-4\right)^2-\left(2x+1\right)^2=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)=\left(5x-5\right)\left(9x-3\right)\)
Bài 2 :
a ) \(3x^3-6x^2+3x^2y-6xy=3x\left(x^2-2x+xy-2y\right)\)
b ) \(x^2-2x+xy-2y=x\left(x-2\right)+y\left(x-2\right)=\left(x-2\right)\left(x+y\right)\)
c ) \(2x+x^2-2y-2xy=......................\)
d ) \(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
e ) \(x^2+y^2-2xy-4=\left(x-y\right)^2-2^2=\left(x-y-2\right)\left(x-y+2\right)\)
f )\(2xy-x^2-y^2+9=-\left(x-y\right)^2+9=3^2-\left(x-y\right)^2=\left(3-x+y\right)\left(3+x-y\right)\)
a) \(2x\left(2x+5\right)-4x\left(x-3\right)=7\)
\(4x^2+10x-4x^2+12x=7\)
\(22x=7\Rightarrow x=0,31\)
b) \(\left(x+2\right)\left(x-2\right)-\left(x+1\right)^2=2\)
\(\left(x^2-4\right)-\left(x^2+2x+1\right)=2\)
\(x^2-4-x^2-2x-1=2\)
\(-2x=7\Rightarrow x=-3,5\)
c) \(\left(x+2\right)\left(x-1\right)-\left(x+3\right)\left(x-2\right)=0\)
\(x^2-x+2x-2-x^2+2x+3x-6=0\)
\(6x=8\Rightarrow x=1,3\)
Lần sau đăng thì chia thành nhiều câu hỏi nhé
\(16^2-9.\left(x+1\right)^2=0\)
\(16^2-\text{ }\left[3.\left(x+1\right)\right]^2=0\)
\(\left[16-3.\left(x+1\right)\right].\left[16+3\left(x+1\right)\right]=0\)
\(\left[16-3x-3\right]\left[16+3x+3\right]=0\)
\(\left[13-3x\right].\left[19+3x\right]=0\)
\(\Rightarrow\orbr{\begin{cases}13-3x=0\\19+3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=13\\3x=-19\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{13}{3}\\x=-\frac{19}{3}\end{cases}}}\)
KL:..............................
a) \(x^2\left(x-3\right)+12-4x=0\)
\(x^2\left(x-3\right)-4\left(x-3\right)=0\)
\(\left(x-3\right)\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x^2-4=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x\in\left\{\pm2\right\}\end{cases}}\)
b) \(x\left(2x-7\right)-3\left(7-2x\right)=0\)
\(x\left(2x-7\right)+3\left(2x-7\right)=0\)
\(\left(2x-7\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-7=0\\x+3=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-3\end{cases}}\)
c) \(\left(2x-1\right)^2-25=0\)
\(\left(2x-1\right)^2-5^2=0\)
\(\left(2x-1-5\right)\left(2x-1+5\right)=0\)
\(\left(2x-6\right)\left(2x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-6=0\\2x+4=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
d) \(\left(3x-5\right)^2-\left(2x-3\right)^2=0\)
\(\left(3x-5-2x+3\right)\left(3x-5+2x-3\right)=0\)
\(\left(x-2\right)\left(5x-8\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\5x-8=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{8}{5}\end{cases}}\)
a, 2x(x-3)-2x2=12
⇔2x2-6x-2x2=12
⇔-6x=12
⇔x=-2
b,(x-2)2 -x(x+3)= 25
⇔(x-2)2 -25-x(x+3)=0
⇔[(x-2)2-52]-x(x+3)=0
⇔(x+3)(x-7)-x(x+3)=0
⇔(x+3)(x-7-x)=0
⇔(x+3)(-7)=0
⇔x+3=0
⇔x=-3
c, 2x (x-3) +4(3-x)=0
⇔ 2x (x-3) -4(x-3)=0
⇔(x-3)(2x-4)=0
⇒\(\left[{}\begin{matrix}x-3=0\\2x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
d,x2 -9x -10= 0
⇔x2 -10x+x-10=0
⇔x(x-10) + (x-10)=0
⇔(x-10)(x+1)=0
\(\Rightarrow\left[{}\begin{matrix}x-10=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-1\end{matrix}\right.\)