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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(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) \(x^3-5x^2+x-5=0\)
\(\Leftrightarrow\left(x^3-5x^2\right)+\left(x-5\right)=0\)
\(\Leftrightarrow x^2\left(x-5\right)+\left(x-5\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vôlí\right)\\x=5\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x\left(x-5\right)-4x+20=0\)
\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-4=0\\x-5=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=4\\x=5\end{array}\right.\)
b) \(x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+6=0\\x-7=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-6\\x=7\end{array}\right.\)
d) \(x^2-9x+8=0\)
\(\Leftrightarrow x^2-x-8x+8=0\)
\(\Leftrightarrow x\left(x-1\right)-8\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-8\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-8=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=8\end{array}\right.\)
g) \(3x^2-5x+2=0\)
\(\Leftrightarrow3x^2-3x-2x+2=0\)
\(\Leftrightarrow3x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\3x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=\frac{2}{3}\end{array}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^3+6x^2-13x-42=0\)
\(\Leftrightarrow\left(x^3-3x^2\right)+\left(9x^2-27x\right)+\left(14x-42\right)=0\)
\(\Leftrightarrow x^2\left(x-3\right)+9x\left(x-3\right)+14\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)+\left(x^2+9x+14\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+7x+2x+14\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[x\left(x+7\right)+2\left(x+7\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)\left(x+7\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-3=0\\x+2=0\\x+7=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=-2\\x=-7\end{array}\right.\)
x3 + 6x2 - 13x - 42 = 0
=> x3 - 3x2 + 9x2 - 27x + 14x - 42 = 0
=> x2 ( x - 3 ) + 9x ( x - 3 ) + 14 ( x - 3 ) = 0
=> ( x - 3 ) ( x2 + 9x + 14) = 0
=> ( x - 3 ) ( x2 + 2x + 7x + 14 ) = 0
=> ( x - 3 ) [ x ( x + 2 ) + 7 ( x + 2 ) ] = 0
=> ( x - 3 ) ( x + 2 ) ( x + 7 ) = 0
=> x - 3 = 0 => x = 3
=> x + 2 = 0 => x = -2
=> x + 7 = 0 => x = -7
![](https://rs.olm.vn/images/avt/0.png?1311)
x2+6x-7x-42=0
x(x+6)-7(x+6)=0
(x+6)(x-7)=0
x=-6 hoac x=7
( nho l ike nha)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tìm x, biết:
a) \(x\left(x-5\right)-4x+20=0\)
\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)
Vậy \(x=5\) hoặc \(x=4\)
b) \(x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+6=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)
Vậy \(x=-6\) hoặc \(x=7\)
c) \(x^3-5x^2+x-5=0\)
\(\Leftrightarrow\left(x^3-5x^2\right)+\left(x-5\right)=0\)
\(\Leftrightarrow x^2\left(x-5\right)+\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x^2+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x^2=-1\left(loại\right)\end{matrix}\right.\)
Vậy \(x=5\)
d) \(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow\left(x^4-2x^3\right)+\left(10x^2-20x\right)=0\)
\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x^3+10x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
Vậy \(x=2\) hoặc \(x=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
bạn đã học giải pt bậc 2 chưa có công thức bài nào cũng giải đc
a) x^2+3x=0
<=> x(x+3)=0
<=> x=0 hoặc x+3=0
<=> x=0 hoặc x=-3
S={0;-3}
b) x^2-x-42=0
<=> x^2-7x+6x-42=0
<=> x(x-7)+6(x-7)=0
<=> (x-7)(x+6)=0
<=> x-7=0 hoac x+6=0
<=> x=7,x=-6
c) ,d) tương tự
e) 2x^3+3x^2-x-1=0
<=> 2x^3+x^2+2x^2+x-2x-1=0
<=> x^2(2x+1)+x(2x+1)-(2x+1)=0
<=> (2x+1)(x^2+x-1)=0
<=>2x+1=0 hoặc x^2+x-1=0
<=> x=-1/2 ,x=-1+căn5/2,x=-1-căn5/2
\(-x^2+x+42=0 \)
<=>\(-x^2-6x+7x+42=0 \)
<=>\(-x\left(x+6\right)+7\left(x+6\right)=0 \)
<=>\(\left(x+6\right)\left(7-x\right)=0 \)
<=>\(\orbr{\begin{cases}x+6=0\\7-x=0\end{cases}}< =>\orbr{\begin{cases}x=-6\\x=7\end{cases}}\)