Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(5\left(x+7\right)-12x=15\)
\(5x+35-12x=15\)
\(-7x=15-35\)
\(-7x=-20\)
\(x=\frac{20}{7}\)
vay \(x=\frac{20}{7}\)
b) \(x^2-25-\left(x+5\right)=0\)
\(x^2-5^2-\left(x+5\right)=0\)
\(\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\)
\(\left(x+5\right)\left(x-5-1\right)=0\)
\(\left(x+5\right)\left(x-6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\x-6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-5\\x=6\end{cases}}\)
vay \(\orbr{\begin{cases}x=-5\\x=6\end{cases}}\)
c) \(\left(2x-1\right)^2-\left(4x^2-1\right)=0\)
\(\left(2x-1\right)\left(2x-1\right)-\left(\left(2x\right)^2-1^2\right)=0\)
\(\left(2x-1\right)\left(2x-1\right)-\left(2x-1\right)\left(2x+1\right)=0\)
\(\left(2x-1\right)\left(2x-1-2x-1\right)=0\)
\(-2.\left(2x-1\right)=0\)
\(\Rightarrow2x-1=0\)
\(\Rightarrow x=\frac{1}{2}\)
vay \(x=\frac{1}{2}\)
d) \(x^2.\left(x^2+4\right)-x^2-4=0\)
\(x^2\left(x^2+4\right)-\left(x^2+4\right)=0\)
\(\left(x^2-1\right)\left(x^2+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2-1=0\\x^2+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=1\\x^2=-4\end{cases}}\Rightarrow\orbr{\begin{cases}x=1hoacx=-1\\kotontai\end{cases}}\)
vay \(x=1\)hoac \(x=-1\)
\(2x^2-7x+5=0\)
\(2x^2-2x-5x+5=0\)
\(2x\left(x-1\right)-5\left(x-1\right)=0\)
\(\left(x-1\right)\left(2x-5\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\2x-5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\2x=5\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\x=\frac{5}{2}\end{array}\right.\)
\(x\left(2x-5\right)-4x+10=0\)
\(x\left(2x-5\right)-2\left(2x-5\right)=0\)
\(\left(2x-5\right)\left(x-2\right)=0\)
\(\left[\begin{array}{nghiempt}x-2=0\\2x-5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=2\\2x=5\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=2\\x=\frac{5}{2}\end{array}\right.\)
\(\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\)
\(x^2-25-x^2+2x=15\)
\(2x=15+25\)
\(2x=40\)
\(x=\frac{40}{2}\)
\(x=20\)
\(x^2\left(2x-3\right)-12+8x=0\)
\(x^2\left(2x-3\right)+4\left(2x-3\right)=0\)
\(\left(2x-3\right)\left(x^2+4\right)=0\)
\(2x-3=0\) (vì \(x^2\ge0\Rightarrow x^2+4\ge4>0\))
\(2x=3\)
\(x=\frac{3}{2}\)
\(x\left(x-1\right)+5x-5=0\)
\(x\left(x-1\right)+5\left(x-1\right)=0\)
\(\left(x-1\right)\left(x+5\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\x+5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\x=-5\end{array}\right.\)
\(\left(2x-3\right)^2-4x\left(x-1\right)=5\)
\(4x^2-12x+9-4x^2+4x=5\)
\(-8x=5-9\)
\(-8x=-4\)
\(x=\frac{4}{8}\)
\(x=\frac{1}{2}\)
\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(5x-2x^2+2x^2-2x=13\)
\(3x=13\)
\(x=\frac{13}{3}\)
\(2\left(x+5\right)\left(2x-5\right)+\left(x-1\right)\left(5-2x\right)=0\)
\(\left(2x+10\right)\left(2x-5\right)-\left(x-1\right)\left(2x-5\right)=0\)
\(\left(2x-5\right)\left(2x+10-x+1\right)=0\)
\(\left(2x-5\right)\left(x+11\right)=0\)
\(\left[\begin{array}{nghiempt}2x-5=0\\x+11=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}2x=5\\x=-11\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-11\end{array}\right.\)
a) \(\left(2x+1\right)^2-4\left(x+2\right)^2=9\)
\(\left(2x+1\right)^2-\left[2\left(x+2\right)\right]^2=9\)
\(\left[2x+1-2\left(x+2\right)\right]\left[2x+1+2\left(x+2\right)\right]=9\)
\(\left(2x+1-2x-4\right)\left(2x+1+2x+4\right)=9\)
\(-3\left(4x+5\right)=9\)
\(4x+5=-3\)
\(4x=-8\)
\(x=-2\)
b) \(x^2-2x-15=0\)
\(x^2-5x+3x-15=0\)
\(x\left(x-5\right)+3\left(x-5\right)=0\)
\(\left(x-5\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}}\)
c) \(2x^2+3x-5=0\)
\(2x^2-2x+5x-5=0\)
\(2x\left(x-1\right)+5\left(x-1\right)=0\)
\(\left(x-1\right)\left(2x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-5}{2}\end{cases}}}\)
a)\(x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
b)\(x^5-9x=0\)
\(\Leftrightarrow x\left(x^4-9\right)=0\)
\(\Leftrightarrow x\left(x^2+3\right)\left(x^2-3\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\x^2+3=0\\x^2-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x\in\varnothing\\x=\pm\sqrt{3}\end{matrix}\right.\)
c)\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
\(\Leftrightarrow x^2\left(x-4\right)-\left(x-4\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(x-4\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+1=0\\x-1=0\\x-4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\x=1\\x=4\end{matrix}\right.\)
d)\(\left(4x^2-25\right)^2-9\left(2x-5\right)^2=0\)
\(\Leftrightarrow\left(2x+5\right)^2\left(2x-5\right)^2-3\left(2x-5\right)^2=0\)
\(\Leftrightarrow\left(2x-5\right)^2\left(4x^2+10x+5-3\right)=0\)
\(\Leftrightarrow\left(2x-5\right)^2\left(4x^2+4x+2x+2\right)=0\)
\(\Leftrightarrow\left(2x-5\right)^2\left[4x\left(x+1\right)+2\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(2x-5\right)^2.2\left(2x+1\right)\left(x+1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}2x-5=0\\2x+1=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}=2,5\\x=-\dfrac{1}{2}=-0,5\\x=-1\end{matrix}\right.\)
a: \(\Leftrightarrow\left(x-2\right)\left(x^2+2x+7+2x+4-5\right)=0\)
=>(x-2)(x^2+4x+6)=0
=>x-2=0
=>x=2
b: =>(2x-5)(2x+5)-(2x-5)(2x+7)=0
=>(2x-5)(2x+5-2x-7)=0
=>2x-5=0
=>x=5/2
c: =>(x+3)(x^2-3x+9+x-9)=0
=>(x+3)(x^2-2x)=0
=>\(x\in\left\{0;2;-3\right\}\)
Ta có:
a) \(x^2-49=0\Leftrightarrow x^2=49\)
\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)
b) \(2x+3-5\left(x+3\right)=0\Leftrightarrow2x+3-5x-15=0\)
\(\Leftrightarrow-3x-12=0\Leftrightarrow x=-4\)
c) \(x^2-2x-15=0\Leftrightarrow\left(x-1\right)^2-4^2=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}\)
a) \(x^2-49=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{49}=7\\-\sqrt{49}=-7\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)
b) 2x + 3 - 5(x + 3) = 0
<=> 2x + 3 - 5x = 0 - 3
<=> 2x - 15 = -3
<=> 2x = -3 + 15
<=> 2x = -12
<=> x = -4
Mong mọi người giúp đỡ mình với!!!
\(2.\left(x-4\right)-x+3=0\)
\(2x-8-x+3=0\)
\(x-5=0\)
\(x=5\)
\(x^2-25-x-5=0\)
\(\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\)
\(\left(x+5\right)\left(x-5-1\right)=0\)
\(\left(x+5\right)\left(x-6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=6\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=-5\\x=6\end{cases}}\)