4, x^2-10x+8x-80=0

x(x-8)+10(x-8)=0

x+10=0         =)x=-10

hoặc

x-8=0        =)x=8

1, =(x+2)(x-2)=0

x+2=0     =)x=-2

hoặc

x-2=0      =)x=2

2,3(x^2-5^2)=0

=x+5=0    =)x=-5

hoặc 

 x-5=0   =)x=5

3,(3+2)^2=25

5^2=25

5, x^2-x-11x+11=0

x(x-1)-11(x-1)=0

x-11=0    =)x=11

hoặc 

x-1=0    =)x=1

xl nheee mk làm nhầm câu 4 trc

là mũ hay nhân vậy ?

bn có thể viết đề rõ hơn k nhìn có vẻ rối

x² - 4 = 0

x² = 4

\(\orbr{\begin{cases}x^2=2^2\\x^2=\left(-2\right)^2\end{cases}}\)

\(\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

3x² - 75 = 0

3x² = 75

x² = 25

\(\orbr{\begin{cases}x^2=5^2\\x^2=\left(-5\right)^2\end{cases}}\)

\(\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)

( x + 2 )² = 25

\(\orbr{\begin{cases}\left(x+2\right)^2=5^2\\\left(x+2\right)^2=\left(-5\right)^2\end{cases}}\)

\(\orbr{\begin{cases}x+2=5\\x+2=-5\end{cases}}\)

\(\orbr{\begin{cases}x=3\\x=-7\end{cases}}\)

ở oooo

e: \(\Leftrightarrow3x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

bạn cứ tra gg rồi ấn thừa số là ra

kinh nghiệm đó

1000%

Câu 9:

\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=-1\end{matrix}\right.\)

\(9,\Leftrightarrow x^2\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow x^2+5x-x-5=0\\ \Leftrightarrow\left(x+5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\\ 12,\Leftrightarrow\left(x+1\right)^2-36=0\\ \Leftrightarrow\left(x+7\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\\ 13,\Leftrightarrow x^3-25x-x^3-8=17\\ \Leftrightarrow-25x=25\Leftrightarrow x=-1\\ 14,\Leftrightarrow x\left(2x^2+8x-3x-12\right)=0\\ \Leftrightarrow x\left(x+4\right)\left(2x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\\x=\dfrac{3}{2}\end{matrix}\right.\)

\(9,x^3-2x^2-x+2=0\\ \Rightarrow x^2\left(x-2\right)-\left(x-2\right)=0\\ \Rightarrow\left(x^2-1\right)\left(x-2\right)=0\\ \Rightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\end{matrix}\right.\)

\(10,\) giống 9

\(11,x^2+4x-5=0\\ \Rightarrow\left(x^2-x\right)+\left(5x-5\right)=0\\ \Rightarrow x\left(x-1\right)+5\left(x-1\right)=0\\ \Rightarrow\left(x-1\right)\left(x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

\(12,2x^2+4x+2=72\\ \Rightarrow2x^2+4x-70=0\\ \Rightarrow x^2+2x-35=0\\ \Rightarrow\left(x^2-5x\right)+\left(7x-35\right)=0\\ \Rightarrow x\left(x-5\right)+7\left(x-5\right)=0\\ \Rightarrow\left(x-5\right)\left(x+7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)

\(13,x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\\ \Rightarrow x\left(x^2-25\right)-\left(x^3+8\right)=17\\ \Rightarrow x^3-25x-x^3-8=17\\ \Rightarrow-25x=25\\ \Rightarrow x=-1\)

\(14,2x^3+5x^2-12x=0\\ \Rightarrow x\left(2x^2+5x-12\right)=0\\ \Rightarrow x\left[\left(2x^2+8x\right)-\left(3x+12\right)\right]=0\\ \Rightarrow x\left[2x\left(x+4\right)-3\left(x+4\right)\right]=0\\ \Rightarrow x\left(2x-3\right)\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\\x=-4\end{matrix}\right.\)

Bài 1:

a) \(\Rightarrow3x^2+3x-2x^2-4x+x+1=0\)

\(\Rightarrow x^2=-1\left(VLý\right)\Rightarrow S=\varnothing\)

b) \(\Rightarrow\left(x-2020\right)\left(2x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2}\end{matrix}\right.\)

c) \(\Rightarrow\left(x-10\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)

d) \(\Rightarrow\left(x+4\right)^2=0\Rightarrow x=-4\)

e) \(\Rightarrow\left(x+6\right)\left(x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)

f) \(\Rightarrow\left(5x-4\right)\left(5x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)

Bài 2:

a) \(\Rightarrow3x\left(x^2-4\right)=0\Rightarrow3x\left(x-2\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

b) \(\Rightarrow x\left(x-2\right)+5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

a)x³-x²=0

⇔x²(x-1)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

b)3x²-5x=0

⇔ x(3x-5)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{3}\end{matrix}\right.\)

c)x³=x⁵

⇔ x³(1-x²)=0

⇔ x³(1-x)(1+x)=0

⇔\(\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

d)(2x+7)²-4(2x+7)=0

⇔ (2x+7)(2x+3)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-7}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)

a) Ta có: \(x^3-x^2=0\)

\(\Leftrightarrow x^2\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

b) Ta có: \(3x^2-5x=0\)

\(\Leftrightarrow x\left(3x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{3}\end{matrix}\right.\)

c) Ta có: \(x^3=x^5\)

\(\Leftrightarrow x^5-x^3=0\)

\(\Leftrightarrow x^3\left(x^2-1\right)=0\)

\(\Leftrightarrow x^3\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

d) Ta có: \(\left(2x+7\right)^2-4\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x+7\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-7}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)

\(x^2+3y^2-4x+6y+7=0\\ \Leftrightarrow\left(x^2-4x+4\right)+\left(3y^2+6y+3\right)=0\\ \Leftrightarrow\left(x-2\right)^2+3\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)

\(3x^2+y^2+10x-2xy+26=0\\ \Leftrightarrow\left(x^2-2xy+y^2\right)+\left(2x^2+10x+\dfrac{25}{8}\right)+\dfrac{183}{8}=0\\ \Leftrightarrow\left(x-y\right)^2+2\left(x^2+2\cdot\dfrac{5}{2}x+\dfrac{25}{4}\right)+\dfrac{183}{8}=0\\ \Leftrightarrow\left(x-y\right)^2+2\left(x+\dfrac{5}{2}\right)^2+\dfrac{183}{8}=0\\ \Leftrightarrow x,y\in\varnothing\)

Sửa đề: \(3x^2+6y^2-12x-20y+40=0\)

\(\Leftrightarrow\left(3x^2-12x+12\right)+\left(6y^2-20y+\dfrac{50}{3}\right)+\dfrac{34}{3}=0\\ \Leftrightarrow3\left(x-2\right)^2+6\left(y^2-2\cdot\dfrac{5}{3}y+\dfrac{25}{9}\right)+\dfrac{34}{3}=0\\ \Leftrightarrow3\left(x-2\right)^2+6\left(y-\dfrac{5}{3}\right)^2+\dfrac{34}{3}=0\\ \Leftrightarrow x,y\in\varnothing\)

\(2\left(x^2+y^2\right)=\left(x+y\right)^2\\ \Leftrightarrow2x^2+2y^2=x^2+2xy+y^2\\ \Leftrightarrow x^2-2xy+y^2=0\\ \Leftrightarrow\left(x-y\right)^2=0\Leftrightarrow x-y=0\Leftrightarrow x=y\)

xy là x.y hay là x và y vậy bn

X và y là số nguyên phải ko