B1: Tìm x
a) (x-3)^2+(4-x)(x+4)=10
b) x^2-2x=0
c) (x^2-9)^2-(x-3)^2=0
Bài 3: Tìm x
a) (2x+3)2−4x2=10
b) (x+1)2−(2+x)(x−2)=0
c) (5x−1)(1+5x)=25x2−7x+15
d) (4−x)2−16=0
e) 3x2−12x=0
g) x2−8x−3x+24=0
e: \(\Leftrightarrow3x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
bài 1 tìm x
a)6x^2-72x=0
b)-2x^4+16x=0
c)x(x-5)-(x-3)^2=0
d)(x-2)^3-(x-2)(x^2+2x+4)=0
a) \(6x^2-72x=0\)
\(6x\left(x-12\right)=0\)
\(6x=0\) hoặc \(x-72=0\)
*) \(6x=0\)
\(x=0\)
*) \(x-12=0\)
\(x=12\)
Vậy \(x=0;x=12\)
b) \(-2x^4+16x=0\)
\(-2x\left(x^3-8\right)=0\)
\(-2x=0\) hoặc \(x^3-8=0\)
*) \(-2x=0\)
\(x=0\)
*) \(x^3-8=0\)
\(x^3=8\)
\(x=2\)
Vậy \(x=0;x=2\)
c) \(x\left(x-5\right)-\left(x-3\right)^2=0\)
\(x^2-5x-x^2+6x-9=0\)
\(x-9=0\)
\(x=9\)
d) \(\left(x-2\right)^3-\left(x-2\right)\left(x^2+2x+4\right)=0\)
\(x^3-6x^2+12x-8-x^3+8=0\)
\(-6x^2+12x=0\)
\(-6x\left(x-2\right)=0\)
\(-6x=0\) hoặc \(x-2=0\)
*) \(-6x=0\)
\(x=0\)
*) \(x-2=0\)
\(x=2\)
Vậy \(x=0;x=2\)
Tìm x
a. 4(x-3)^2-(2x-1)(2x+1)=10
b. x^3-25x=0
\(a,\Leftrightarrow4x^2-24x+36-4x^2+1=10\\ \Leftrightarrow-24x=-27\Leftrightarrow x=\dfrac{9}{8}\\ b,\Leftrightarrow x\left(x^2-25\right)=0\\ \Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
\(a,4.\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
\(\Leftrightarrow4.\left(x^2-6x+9\right)-\left(2x^2\right)-1^2=10\)
\(\Leftrightarrow4x^2-24x+36-4x^2+1=10\)
\(\Leftrightarrow-24x+27=10\)
\(\Leftrightarrow-24x=-27\)
\(\Leftrightarrow x=\dfrac{27}{24}\)
Vậy \(x=\dfrac{27}{24}\)
\(b,x^3-25x=0\)
\(\Leftrightarrow x\left(x^2-25\right)=0\)
\(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
Vậy \(x\in\left\{0;\pm5\right\}\)
tìm x
a,(x+1)^3-(x-1)^3-6(x-1)^2=-10
b,x(x+5)(x-5)-(x+2)(x^2-2x+4)=42
c,(x-2)^3-(x-3)(x^2+3x+9)+6(x+1)^2=49
a) \(\left(x+1\right)^3-\left(x-1\right)^3-6\cdot\left(x-1\right)^2=10\)
\(\Rightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\cdot\left(x^2-2x+1\right)=10\)
\(\Rightarrow6x^2+2-6x^2+12x-6=10\)
\(\Rightarrow12x-4=10\)
\(\Rightarrow12x=14\)
\(\Rightarrow x=\dfrac{7}{6}\)
b) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=42\)
\(\Rightarrow x\left(x^2-25\right)-\left(x^3+8\right)=42\)
\(\Rightarrow x^3-25x-x^3-8=42\)
\(\Rightarrow-25x-8=42\)
\(\Rightarrow-25x=50\)
\(\Rightarrow x=\dfrac{50}{-25}=-2\)
c) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
\(\Rightarrow x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=49\)
\(\Rightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)
\(\Rightarrow24x+25=49\)
\(\Rightarrow24x=24\)
\(\Rightarrow x=\dfrac{24}{24}=1\)
tìm x
a) 3.(x-3)-4x+12=0
b)(x+2)^2-(x+2).(x-2) =0
c)x^3+3x=3x^2+1
d)2/3x.(x^2-4)=0
e)(2x-3)^2-(+5)^2=0
\(a,=3x-9-4x+12=-x+3=0\)
\(\Leftrightarrow x=3\)
Vậy ..
\(b,=\left(x+2\right)\left(x+2-x+2\right)=4\left(x+2\right)=0\)
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
Vậy ..
\(c,=x^3-3x^2+3x-1=\left(x-1\right)^3=0\)
\(\Leftrightarrow x=1\)
Vậy ..
\(d,\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy ..
\(e,=\left(2x-3-5\right)\left(2x-3+5\right)=\left(2x-8\right)\left(2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{2}=4\\x=-\dfrac{2}{2}=-1\end{matrix}\right.\)
Vậy ...
a) Ta có: 3(x-3)-4x+12=0
\(\Leftrightarrow3\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow x-3=0\)
hay x=3
Vậy: S={3}
b) Ta có: \(\left(x+2\right)^2-\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow x^2+4x+4-x^2+4=0\)
\(\Leftrightarrow4x=-8\)
hay x=-2
Vậy: S={-2}
c) Ta có: \(x^3+3x=3x^2+1\)
\(\Leftrightarrow x^3-3x^2+3x-1=0\)
\(\Leftrightarrow x-1=0\)
hay x=1
Vậy: S={1}
d) Ta có: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: S={0;2;-2}
a) 3.(x-3)-4x+12=0
=> 3x - 9 - 4x + 12 = 0
=> -x + 3 = 0
=> x = 3
b) (x+2)^2-(x+2).(x-2) =0
\(\Rightarrow\left(x+2\right)^2-x^2+4=0\)
\(\Rightarrow x^2+4x+4-x^2+4=0\)
=> 4x + 8 = 0
=> x = -2
c) x^3+3x=3x^2+1
\(\Rightarrow x^3+3x-3x^2-1=0\)
\(\Rightarrow\left(x-1\right)^3=0\)
=> x = 1
d) \(\dfrac{2}{3}x\left(x^2-4\right)=0\)
\(\Rightarrow\dfrac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)
=> x = 0 hoặc x = 2 hoặc x = -2
e) \(\left(2x-3\right)^2-5^2=0\)
\(\Rightarrow\left(2x-8\right)\left(2x+2\right)=0\)
=> x = 4 hoăc x = -1
Tìm x
a, 3/4x*(x2-9)=0
b, x3-16x=0
c, (x-1)(x+2)-x-2=0
d, 3x3-27x=0
e, x2(x+1)+2x(x+1)=0
f, x(2x-3)-2(3-2x)=0
c: =>(x-1)(x+1)=0
hay \(x\in\left\{1;-1\right\}\)
a,
\(=\dfrac{3}{4x}.\left(x-3\right)\left(x+3\right)\)=0
\(\left\{{}\begin{matrix}\dfrac{3}{4x}=0\\x-3=0\\x+3=0\end{matrix}\right.\)
=>\(x=\left\{3,-3\right\}\)
b,
\(x^3-16x=0\\x\left(x^2-16\right)\\ x\left(x-4\right)\left(x+4\right)\)
\(\left\{{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)
=>\(x=\left\{-4,0,4\right\}\)
d,
\(3x^3-27x=0\\ 3x\left(x^2-9\right)=0\\ 3x\left(x-3\right)\left(x+3\right)=0\)
\(\left\{{}\begin{matrix}3x=0\\x-3=0\\x+3=0\end{matrix}\right.\)
=>\(x=\left\{-3,0,3\right\}\)
e,
\(x^2+\left(x+1\right)+2x\left(x+1\right)=0\\ x\left(x+1\right)\left(x+2\right)=0\)
\(\left\{{}\begin{matrix}x=0\\x+1=0\\x+2=0\end{matrix}\right.\)
=>\(x=\left\{-2,-1,0\right\}\)
f,
\(x\left(2x-3\right)-2\left(3-2x\right)=0\\ \left(2x-3\right)\left(x+2\right)=0\)
\(\left\{{}\begin{matrix}2x-3=0\\x+2=0\end{matrix}\right.\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)
tìm x biết
a) (2x-3)(2x+3)=0
b) x^2-1=0
c) x^2-9=0
d) 4^2-16=0
e) 25x^2-9=0
a) \(\left(2x-3\right)\left(2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b) \(x^2-1=0\Rightarrow\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
c) \(x^2-9=0\Rightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
d) \(\Rightarrow\left(2x-4\right)\left(2x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
2) \(\Rightarrow\left(5x-3\right)\left(5x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
Tìm x biết:
a, 16x² – 9(x + 1)²= 0
b, x2 (x – 1) – 4x2 + 8x – 4 = 0
c, x(2x – 3) – 2(3 – 2x) = 0
d, (x – 3)(x² + 3x + 9) – x(x + 2)(x – 2) = 1
e, 4x² + 4x – 6 = 2
f, 2x² + 7x + 3 = 0
e: ta có: \(4x^2+4x-6=2\)
\(\Leftrightarrow4x^2+4x-8=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
f: Ta có: \(2x^2+7x+3=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
tìm x
a, (2x - 3)\(^2\) = |3 - 2x|
b, (x - 1)\(^2\) + (2x - 1)\(^2\) - 0
c, x - 2\(\sqrt{x}\) = 0
d, (x - 1)\(^2\) + 1/7 = 0
a: \(\left(2x-3\right)^2=\left|3-2x\right|\)
=>\(\left\{{}\begin{matrix}\left|2x-3\right|>=0\\\left(2x-3\right)^2=\left(2x-3\right)\end{matrix}\right.\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)=0\)
=>\(\left(2x-3\right)\left(2x-3-1\right)=0\)
=>\(\left(2x-3\right)\left(2x-4\right)=0\)
=>\(\left[{}\begin{matrix}2x-3=0\\2x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\)
b: \(\left(x-1\right)^2+\left(2x-1\right)^2=0\)
=>\(x^2-2x+1+4x^2-4x+1=0\)
=>\(5x^2-6x+2=0\)
\(\Delta=\left(-6\right)^2-4\cdot5\cdot2=36-20\cdot2=-4< 0\)
=>Phương trình vô nghiệm
c: ĐKXĐ: x>=0
\(x-2\sqrt{x}=0\)
=>\(\sqrt{x}\cdot\sqrt{x}-2\cdot\sqrt{x}=0\)
=>\(\sqrt{x}\left(\sqrt{x}-2\right)=0\)
=>\(\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=4\left(nhận\right)\end{matrix}\right.\)
d: \(\left(x-1\right)^2+\dfrac{1}{7}=0\)
mà \(\left(x-1\right)^2+\dfrac{1}{7}>=\dfrac{1}{7}>0\forall x\)
nên \(x\in\varnothing\)