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a: \(\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\)
=>-12x-15=9
=>-12x=24
hay x=-2
b: \(\Leftrightarrow9x^2-6x+1+2x^2+12x+18+11\left(1-x^2\right)=6\)
\(\Leftrightarrow11x^2+6x+19+11-11x^2=6\)
=>6x+30=6
=>6x=-24
hay x=-4
c: \(\Leftrightarrow x^3+3x^2+3x+1-x^3-3x^2=2\)
=>3x=1
hay x=1/3
d: \(\Leftrightarrow x^3-6x^2+12x-8-x\left(x^2-1\right)+6x^2=5\)
\(\Leftrightarrow x^3+12x-8-x^3+x=5\)
=>13x=13
hay x=1
e: \(\Leftrightarrow x^3-27-x^3+16x=5\)
=>16x=32
hay x=2
\(x\left(x-3\right)+x-3=0\)
\(\left(x-3\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}}\)
KL:......................
\(x^3-5x=0\)
\(x\left(x^2-5\right)=0\)
Làm tương tự như câu a
@_@ n...h..i......ề....u q...u.....................á!
a) Ta có: (2x-3)(x+2)=0
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{3}{2};-2\right\}\)
b) Ta có: (3x-1)(2x-5)=(3x-1)(x+2)
⇔\(\left(3x-1\right)\left(2x-5\right)-\left(3x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left[\left(2x-5\right)-\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(3x-1\right)\left(2x-5-x-2\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=7\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{3};7\right\}\)
c) Ta có: \(\left(x^2-25\right)+\left(x-5\right)\left(2x-11\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)+\left(x-5\right)\left(2x-11\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+5+2x-11\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3x-6\right)=0\)
\(\Leftrightarrow\left(x-5\right)\cdot3\cdot\left(x-2\right)=0\)
mà 3≠0
nên \(\left[{}\begin{matrix}x-5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)
Vậy: x∈{5;2}
d) Ta có: \(\left(x^2-6x+9\right)-4=0\)
\(\Leftrightarrow\left(x-3\right)^2-2^2=0\)
\(\Leftrightarrow\left(x-3-2\right)\left(x-3+2\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
Vậy: x∈{5;1}
e) Ta có: \(2x^3-5x^2+3x=0\)
\(\Leftrightarrow x\left(2x^2-5x+3\right)=0\)
\(\Leftrightarrow x\left(2x^2-2x-3x+3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x-1\right)-3\left(x-1\right)\right]=0\)
\(\Leftrightarrow x\left(x-1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;1;\frac{3}{2}\right\}\)
giải
a)(x+2)^2+(x-3)^2-2(x-1)(x+1)
x^2+4x+4+x^2-6x+9-2(x^2-1)=9
x^2+4x+4+x^2-6x+9-2x^2+2=9
-2x+15=9
-2x=9-15
-2x=-6
x=-6:(-2)
x=3
d)4(x+1)^2+(2x-1)^2-8(x-1)(x+1)=11
4(x^2+2x+1)+4x^2+4x+1-8(x^2-1)
4x^2+8x+4+4x^2+4x+1-8x^2+8=11
12x+13=11
12x=11-13
12x=-2
x=-1/6
a. x ( x + 4 ) ( 4 - x ) + ( x - 5 ) ( x2 + 5x + 25 ) = 3
- x ( x + 4 ) ( x - 4) + x3 - 53 = 3
-x . (x2 - 42) + x3 -125 = 3
-x3 + 16x + x3 - 125 = 3
16x - 125 = 3
16x = 128
x =8
a, \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x+4\right)\left(x-4\right)=5\)
\(\Rightarrow x^3+3x^2+9x-3x^2-9x-27-x\left(x^2-16\right)=5\)
\(\Rightarrow x^3-27-x^3-16x=5\)
\(\Rightarrow-16x-27=5\)
\(\Rightarrow-16x=32\Rightarrow x=-2\)
b, \(\left(x-2\right)^3-\left(x+5\right)\left(x^2-5x+25\right)+6x^2=11\)
\(\Rightarrow x^3-6x^2+12x-8-\left(x^3-5x^2+25x+5x^2-25x+125\right)+6x^2=11\)
\(\Rightarrow x^3-6x^2+12x-8-x^3-125+6x^2=11\)
\(\Rightarrow12x-133=11\Rightarrow12x=144\Rightarrow x=12\)
Chúc bạn học tốt!!!
a)
\(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x+4\right)\left(x-4\right)=5\)
\(\Rightarrow x^3-3^3-x.\left(x^2-16\right)=5\)
\(\Rightarrow x^3-27-x^3+16.x=5\)
\(\Rightarrow16x-27=5\)
\(\Rightarrow16x=32\)
\(\Rightarrow x=2\)
Vậy x = 2
b)
\(\left(x-2\right)^3-\left(x+5\right)\left(x^2-5x+25\right)+6x^2=11\)
\(\Rightarrow x^3-6x^2+12x-8-x^3-125+6x^2=11\)
\(\Rightarrow12x-133=11\)
\(\Rightarrow12x=144\)
\(\Rightarrow x=12\)
Vậy x = 12