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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
<=> 2x + 42 - 2 - x + 42 = 0
<=> x = 0 - 42 + 2 - 42
<=> x = -82
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ (x-5)^2-49=0
<=>(x-5)2-72
<=>(x-5-7)(x-5+7)=0
<=>(x-12)(x+2)=0
<=>x-12=0 hoặc x+2=0
<=>x=12 hoặc x=-2
vậy x=12 hoặc x=-2
b/ (x+11)^2=121
<=>(x+11)2-121=0
<=>(x+11)2-112=0
<=>(x+11-11)(x+11+11)=0
<=>x(x+22)=0
<=>x=0 hoặc x+22=0
<=>x=0 hoặc x=-22
vậy x=0 hoặc x=-22
c/ x.(x+7)-6x-42=0
<=>x2+7x-6x-42=0
<=>x2+x-42=0
<=>x2-6x+7x-42=0
<=>x(x-6)+7(x-6)=0
<=>(x-6)(x-7)=0
<=>x-6=0 hoặc x-7=0
<=>x=6 hoặc x=7
vậy x=6;7
d/ x^4-2x^3+10x^2-20x=0
<=>x3(x-2)+10x(x-2)=0
<=>(x-2)(x3+10x)=0
<=>(x-2)x(x2+10)=0
<=>x-2=0 hoặc x=0 hoặc x2+10=0
<=>x=2 hoặc x=0 hoặc x2=-10(vô lí)
vậy x=2;0
a)(x-5)2-49=0
<=>(x-5-7)(x-5+7)=0
<=>(x-12)(x+2)=0
<=>x-12=0 hoặc x+2=0
<=>x=12 hoặc x=-2
b)(x+11)2=121
<=>(x+11)2-121=0
<=>(x+11-11)(x+11+11)=0
<=>x(x+22)=0
<=>x=0 hoặc x+22=0
<=>x=0 hoặc x=-22
c)x(x+7)-6x-42=0
<=>x(x+7)-(6x+42)=0
<=>x(x+7)-6(x+7)=0
<=>(x+7)(x-6)=0
<=>x+7=0 hoặc x-6=0
<=>x=-7 hoặc x=6
d)x4-2x3+10x2-20x=0
<=>x(x3-2x2+10x-20)=0
<=>x[(x3-2x2)+(10x-20)]=0
<=>x[x2(x-2)+10(x-2)]=0
<=>x(x-2)(x2+10)=0
Do x2>0=>x2+10>0
=>x(x-2)=0
<=>x=0 hoặc x-2=0
<=>x=0 hoặc x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,PT\Leftrightarrow3x^2+3x-2x^2-4x=-1-x\Leftrightarrow x^2=-1\left(\text{vô nghiệm}\right)\)
Vậy: ...
\(b,PT\Leftrightarrow4x\left(x-2019\right)-\left(x-2019\right)=0\Leftrightarrow\left(x-2019\right)\left(4x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2019\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy: ...
\(c,PT\Leftrightarrow\left(x-4-6\right)\left(x-4+6\right)=0\Leftrightarrow\left(x-10\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)
Vậy: ...
\(d,PT\Leftrightarrow\left(x+4\right)^2=0\Leftrightarrow x=-4\)
Vậy: ...
\(e,PT\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)
Vậy: ...
\(f,PT\Leftrightarrow\left(5x-3\right)\left(5x+3\right)=0\Leftrightarrow x=\pm\dfrac{3}{5}\)
Vậy: ...
câu c sao tính ra vậy đc vậy k hiểu giải thích hộ e đi 36 đâu mất òi
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-5x-4\left(x-5\right)=0\)
\(\Leftrightarrow\)\(x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\)\(\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=4\end{cases}}\)
Vậy....
\(2x\left(x+6\right)=7x+42\)
\(\Leftrightarrow\)\(2x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow\)\(2x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\)\(\left(x+6\right)\left(2x-7\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+6=0\\2x-7=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-6\\x=\frac{7}{2}\end{cases}}\)
Vậy......
\(x^3-5x^2+x-5=0\)
\(\Leftrightarrow\)\(x^2\left(x-5\right)+\left(x-5\right)=0\)
\(\Leftrightarrow\)\(\left(x-5\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\)\(x-5=0\)
\(\Leftrightarrow\)\(x=5\)
\(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow\)\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\)\(x\left(x-2\right)\left(x^2+10\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
a: (x^2+9)(9x^2-1)=0
=>9x^2-1=0
=>x^2=1/9
=>x=1/3 hoặc x=-1/3
b: (4x^2-9)(2^(x-1)-1)=0
=>4x^2-9=0 hoặc 2^(x-1)-1=0
=>x^2=9/4 hoặc x-1=0
=>x=1;x=3/2;x=-3/2
c: (3x+2)(9-x^2)=0
=>(3x+2)(3-x)(3+x)=0
=>\(\left[{}\begin{matrix}3x+2=0\\3-x=0\\3+x=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{2}{3};3;-3\right\}\)
d: (3x+3)^2(4x-4^2)=0
=>3x+3=0 hoặc 4x-16=0
=>x=4 hoặc x=-1
e: \(2^{\left(x-5\right)\left(x+2\right)}=1\)
=>(x-5)(x+2)=0
=>x-5=0 hoặc x+2=0
=>x=5 hoặc x=-2
![](https://rs.olm.vn/images/avt/0.png?1311)
a. x(x-5) -4x+20=0
<=> x(x-5) - 4(x-5)=0
<=> (x-5)(x-4)=0
<=>(x-5)=0 hoặc x-4=0
<=> x=5 hoặc x=4
Vậy x={4;5}
b.tương tự
c. x3-5x2+x-5 =0
<=> x2(x-5) + (x-5) = 0
<=> (x-5) (x2+1) = 0
<=> x-5=0 hoặc x2+1=0(loại vì x2=-1)
<=> x=5
vậy x=5
d. bạn kiểm tra lại đề
Tìm x :
a) \(x\left(x-5\right)-4x+20=0\)
\(\Leftrightarrow x^2-5x-4x+20=0\)
\(\Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\)
\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
b) \(x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow x^2+6x-7x-42=0\)
\(\Leftrightarrow\left(x^2+6x\right)-\left(7x+42\right)=0\)
\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-6\end{matrix}\right.\)
c) \(x^3-5x^2+x-5=0\)
\(\Leftrightarrow\left(x^3-5x^2\right)+\left(x-5\right)=0\)
\(\Leftrightarrow x^2\left(x-5\right)+\left(x-5\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vôlí\right)\\x=5\end{matrix}\right.\)
\(-x^2+x+42=0 \)
<=>\(-x^2-6x+7x+42=0 \)
<=>\(-x\left(x+6\right)+7\left(x+6\right)=0 \)
<=>\(\left(x+6\right)\left(7-x\right)=0 \)
<=>\(\orbr{\begin{cases}x+6=0\\7-x=0\end{cases}}< =>\orbr{\begin{cases}x=-6\\x=7\end{cases}}\)