Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)
\(\left(x+2\right)^2-9=0\)
\(\Rightarrow\left(x+2\right)^2=9=3^2\)
\(\Rightarrow x+2=\pm3\)
\(\Rightarrow x=-5;1\)
b)
\(25x^2-10x+1=0\)
\(\left(5x\right)^2-2\cdot5x+1^2=0\)
\(\Rightarrow\left(5x+1\right)^2=0\)
\(\Rightarrow5x+1=0\)
\(\Rightarrow5x=-1;x=\dfrac{-1}{5}\)
c)
\(x^2+14x+49=0\)
\(\Rightarrow x^2+2\cdot7x+7^2=0\)
\(\Rightarrow\left(x+7\right)^2=0;x+7=0\)
\(\Rightarrow x=-7\)
d)
\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(4x^2-4x+1+x^2+6x+9-5x^2+5\cdot49=0\)
\(\Rightarrow5x^2-5x^2-4x+6x+10+245=0\)
\(\Rightarrow2x+255=0\)
\(\Rightarrow2x=-255\)
\(\Rightarrow x=\dfrac{-255}{2}\)
a,\(3x\left(x-1\right)+x-1=0\)
\(\Rightarrow3x\left(x-1\right)+\left(x-1\right)=0\)
\(\Rightarrow\left(3x+1\right).\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x+1=0\\x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=1\end{matrix}\right.\)
c,\(\left(2x-1\right)^2-25=0\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow\left(2x-1\right)^2=5^2\)
\(\Rightarrow2x-1=\pm5\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Bài 4:
a: \(\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2-12=2\)
\(\Leftrightarrow3x-40=2\)
=>3x=42
hay x=14
b: \(\Leftrightarrow x^3+8-x^3-2x=0\)
=>-2x+8=0
=>-2x=-8
hay x=4
c: \(x\left(x-2\right)+\left(x-2\right)=0\)
=>(x-2)(x+1)=0
=>x=2 hoặc x=-1
d: \(5x\left(x-3\right)-x+3=0\)
=>5x(x-3)-(x-3)=0
=>(x-3)(5x-1)=0
=>x=3 hoặc x=1/5
e: \(3x\left(x-5\right)-\left(x-1\right)\left(3x+2\right)=30\)
\(\Leftrightarrow3x^2-15x-3x^2-2x+3x+2=30\)
=>-14x=28
hay x=-2
f: \(\Leftrightarrow\left(x+2\right)\left(x+30-x-5\right)=0\)
=>x+2=0
hay x=-2
Bài 9 : Tìm x, biết :
a, (x - 2)(x - 3) + (x - 2) - 1 = 0
\(\Leftrightarrow\left(x-2\right)\left(x-3+1\right)-1=0\)
\(\Leftrightarrow\left(x-2\right)^2-1=0\)
\(\Leftrightarrow\left(x-2+1\right)\left(x-2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Vậy x ={1; 3}
b, (x + 2)2 - 2x(2x + 3) = (x + 1)2
\(\Leftrightarrow\left(x+2\right)^2-\left(x+1\right)^2-2x\left(2x+3\right)=0\)
\(\Leftrightarrow\left(x+2+x+1\right)\left(x+2-x-1\right)-2x\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3-2x\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(1-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\1-2x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy \(x=\left\{-\frac{3}{2};\frac{1}{2}\right\}\)
c, 6x3 + x2 = 2x
\(\Leftrightarrow6x^3+x^2-2x=0\)
\(\Leftrightarrow x\left(6x^2+x-2\right)=0\)
\(\Leftrightarrow x\left(6x^2+4x-3x-2\right)=0\)
\(\Leftrightarrow x\left[2x\left(3x+2\right)-\left(3x+2\right)\right]=0\)
\(\Leftrightarrow x\left(3x+2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x+2=0\\2x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\frac{2}{3}\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy \(x=\left\{0;-\frac{2}{3};\frac{1}{2}\right\}\)
a) Ta có: \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{\left(2x+1\right)^2\cdot3}{15}-\frac{5\left(x-1\right)^2}{15}-\frac{7x^2-14x-5}{15}=0\)
\(\Leftrightarrow3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)-7x^2+14x+5=0\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\)
\(\Leftrightarrow36x+3=0\)
\(\Leftrightarrow36x=-3\)
\(\Leftrightarrow x=\frac{-3}{36}\)
Vậy: \(x=\frac{-3}{36}\)
b) Ta có: \(\frac{201-x}{99}+\frac{203-x}{97}=\frac{205-x}{95}+3=0\)
\(\Leftrightarrow\frac{201-x}{99}+\frac{203-x}{97}-\frac{205-x}{95}-3=0\)
\(\Leftrightarrow\left(\frac{201-x}{99}+1\right)+\left(\frac{203-x}{97}+1\right)+\left(\frac{205-x}{95}+1\right)=0\)
\(\Leftrightarrow\frac{201-x+99}{99}+\frac{203-x+97}{97}+\frac{205-x+95}{95}=0\)
\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)
\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\ne0\)
nên 300-x=0
\(\Leftrightarrow x=300\)
Vậy: x=300
c) Ta có: \(x^3+x^2+x+1=0\)
\(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)(1)
Ta có: \(x^2\ge0\forall x\)
\(\Rightarrow x^2+1\ge1\ne0\forall x\)(2)
Từ (1) và (2) suy ra x+1=0
hay x=-1
Vậy: x=-1
d) Ta có: \(\left(x-1\right)x\left(x+1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)
Đặt \(x^2+x-1=t\)
\(\Leftrightarrow\left(t+1\right)\left(t-1\right)=24\)
\(\Leftrightarrow t^2-1-24=0\)
\(\Leftrightarrow t^2-25=0\)
\(\Leftrightarrow\left(t-5\right)\left(t+5\right)=0\)
\(\Leftrightarrow\left(x^2+x-1-5\right)\left(x^2+x-1+5\right)=0\)
\(\Leftrightarrow\left(x^2+x-6\right)\left(x^2+x+4\right)=0\)
\(\Leftrightarrow\left(x^2+3x-2x-6\right)\left(x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{15}{4}\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)\left[\left(x+\frac{1}{2}\right)^2+\frac{15}{4}=0\right]\)(3)
Ta có: \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}\ne0\forall x\)(4)
Từ (3) và (4) suy ra
\(\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-3;2\right\}\)
e) Ta có: \(\left(5x-3\right)-\left(4x-7\right)=0\)
\(\Leftrightarrow5x-3-4x+7=0\)
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Vậy: x=-4
f) Ta có: \(3x^2+2x-1=0\)
\(\Leftrightarrow3x^2+3x-x-1=0\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{-1;\frac{1}{3}\right\}\)
g) Ta có: \(x^2+6x-16=0\)
\(\Leftrightarrow x^2-2x+8x-16=0\)
\(\Leftrightarrow x\left(x-2\right)+8\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-8\right\}\)
h) Ta có: \(x^2+3x-10=0\)
\(\Leftrightarrow x^2+5x-2x-10=0\)
\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-5;2\right\}\)
i) Ta có: \(x^2+x-2=0\)
\(\Leftrightarrow x^2-x+2x-2=0\)
\(\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{1;-2\right\}\)
k) Ta có: \(3x^2+7x+2=0\)
\(\Leftrightarrow3x^2+6x+x+2=0\)
\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\3x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\frac{-1}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{-2;\frac{-1}{3}\right\}\)
l) Ta có: \(4x^2-12x+5=0\)
\(\Leftrightarrow4x^2-2x-10x+5=0\)
\(\Leftrightarrow2x\left(2x-1\right)-5\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{5}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{2};\frac{5}{2}\right\}\)
a: =>(x-5)(x+5)+(x-5)(3x-15)=0
=>(x-5)(x+5+3x-15)=0
=>(x-5)(4x-10)=0
=>x=5 hoặc x=5/2
c: =>x^3-3x^2+2x^2-6x-8x+24=0
=>(x-3)(x^2+2x-8)=0
=>(x-3)(x+4)(x-2)=0
=>\(x\in\left\{3;-4;2\right\}\)