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a: \(\Leftrightarrow\left(2x-3\right)^2-5x\left(2x-3\right)=0\)
=>(2x-3)(-3x-3)=0
=>x=-1 hoặc x=3/2
b: \(\Leftrightarrow49\left(x^2-10x+25\right)-8x-4=0\)
=>\(49x^2-498x+1221=0\)
=>\(x\in\left\{6.03;4.13\right\}\)
c: \(\Leftrightarrow\left(x+6\right)\left(x+6-8\right)=0\)
=>(x-2)(x+6)=0
=>x=2 hoặc x=-6
d: =>\(\left(16x+24\right)^2-\left(x-6\right)^2=0\)
=>(16x+24+x-6)(16x+24-x+6)=0
=>(17x+18)(15x+30)=0
=>x=-2 hoặc x=-18/17
a) 2x (x-5) -(x2-10x +25)=0
\(\Leftrightarrow\)2x(x-5)-(x-5)2=0
\(\Leftrightarrow\)(x-5)(2x-x+5)=0
\(\Leftrightarrow\)(x-5)(x+5)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-5=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
b) x2 - 9 +3x(x+3) = 0
\(\Leftrightarrow\)(x2 - 9) +3x(x+3) =0
\(\Leftrightarrow\)(x-3)(x+3)+3x(x+3)=0
\(\Leftrightarrow\)(x+3)(x-3+3x)=0
\(\Leftrightarrow\)(x+3)(4x-3)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x+3=0\\4x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\4x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{3}{4}\end{matrix}\right.\)
c) x3 - 16x = 0
\(\Leftrightarrow\)x(x2-16)=0
\(\Leftrightarrow\)x(x-4)(x+4)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
d) (2x+3)(x-2) - (x2 -4x+4) = 0
\(\Leftrightarrow\)(2x+3)(x-2) -(x-2)2=0
\(\Leftrightarrow\)(x-2)(2x+3-x+2)=0
\(\Leftrightarrow\)(x-2)(x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
e) 9x2 -(x2 -2x +1)=0
\(\Leftrightarrow\)(3x)2-(x-1)2=0
\(\Leftrightarrow\)(3x-x+1)(3x+x-1)=0
\(\Leftrightarrow\)(2x+1)(4x-1)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x+1=0\\4x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=-1\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
f)x3-4x2 -9x +36 = 0
\(\Leftrightarrow\)(x3-9x)-(4x2-36)=0
\(\Leftrightarrow\)x(x2-9)-4(x2-9)=0
\(\Leftrightarrow\)(x-4)(x2-9)=0
\(\Leftrightarrow\)(x-4)(x-3)(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)
g) 3x - 6 = (x-1).(x-2)
\(\Leftrightarrow\)3(x-2)=(x-1)(x-2)
\(\Leftrightarrow\)x-1=3
\(\Leftrightarrow\)x=4
i) (x-2).(x+2) +(2x+1)2 =-5x.(x-3) =5 (?? đề sao vậy ??)
k) x2 -1 = (x-1).(2x+3)
\(\Leftrightarrow\)(x-1)(x+1)=(x-1)(2x+3)
\(\Leftrightarrow\)x+1=2x+3
\(\Leftrightarrow\)x-2x=3-1
\(\Leftrightarrow\)-x=2
\(\Leftrightarrow\)x=-2
l) (2x-1)2 +(x+3).(x-3) -5x(x-2)=6
\(\Leftrightarrow\)4x2-4x+1+x2-9-5x2+10x=6
\(\Leftrightarrow\)6x-8=6
\(\Leftrightarrow\)6x=14
\(\Leftrightarrow\)x=\(\frac{7}{3}\)
a, \(x\left(x-3\right)-x^2+2=0\)
\(\Leftrightarrow x^2-3x-x^2+2=0\\ \Leftrightarrow-3x+2=0\)
\(\Leftrightarrow-3x=-2\\ \Rightarrow x=\frac{2}{3}\)
b, \(x^2-2x+1=0\\ \Leftrightarrow\left(x-1\right)^2=0^2\)
\(\Leftrightarrow x-1=0\\ \Leftrightarrow x=1\)
c, x(x-1)-(x+3)(x+4)=5x
\(\Leftrightarrow x^2-x-x^2-4x-3x-12=5x\)
\(\Leftrightarrow x^2-x-x^2-4x-3x-5x=12\\ \Leftrightarrow-13x=12\\ \Rightarrow x=\frac{-12}{13}\)
d, ko có vế phải ạ
e, \(x^2+2x=15\)
\(\Leftrightarrow\left(x^2+2x+1\right)-16=0\\ \Leftrightarrow\left(x+1\right)^2-4^2=0\)
\(\Leftrightarrow\left(x+1-4\right)\left(x+1+4\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+5\right)=0\)
\(\left[{}\begin{matrix}x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
f, \(x^4-5x^3+4x^2=0\)
\(\Leftrightarrow x^4-x^3-4x^3+4x^2=0\\ \Leftrightarrow x^3\left(x-1\right)-4x^2\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^3-4x^2\right)=0\)
\(\Leftrightarrow\left(x-1\right).x^2\left(x-4\right)=0\)
\(\left[{}\begin{matrix}x^2=0\\x-1=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=4\end{matrix}\right.\)
chắc bn nảy hỏi lun cả bài tâp về nhà quá, làm km 1 câu
a) = a+a+a + a +a +1 -a -a -a = a(a+a+1) +(a+a+1) - a(a+a+1)= (a+a+1)(a-a+1)
tự bn thêm mũ 4;3;2 vào được là bn làm dc cac câu sau
\(a,x^3-16x=0\)
\(\Leftrightarrow x\left(x^2-16\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
\(b,x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)
\(\Leftrightarrow\left(x-2\right)x\left(x^2+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x=0\\x^2+10=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=0\\\left[{}\begin{matrix}x^2=10\\x^2=-10\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=0\\x=\sqrt{10}\\x=-\sqrt{10}\end{matrix}\right.\)\(c,\left(2x-1\right)^2=\left(x+3\right)^2\)
\(\Leftrightarrow4x^2-4x+1=x^2+6x+9\)
\(\Leftrightarrow4x^2-4x+1-x^2-6x-9=0\)
\(\Leftrightarrow3x^2-10x-8=0\)
\(\Leftrightarrow3x^2-12x+2x-8=0\)
\(\Leftrightarrow3x\left(x-4\right)+2\left(x-4\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x-2=0\\x-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=4\end{matrix}\right.\)
Phần d tương tự
Câu a :
\(x^3-16x=0\)
\(\Leftrightarrow x\left(x^2-4^2\right)=0\)
\(\Leftrightarrow x\left[\left(x+4\right)\left(x-4\right)\right]=0\)
\(\Rightarrow\) \(x=0\)
\(\Rightarrow\) \(x+4=0\Rightarrow x=-4\)
\(\Rightarrow x-4=0\Rightarrow x=4\)
Câu b :
\(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)\) \(=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\)
\(\Rightarrow x=0\)
\(\left(x-2\right)=0\Rightarrow x=2\)
\(x^2+10=0\) \(\Rightarrow\) x ( loại )