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a,2x^2=x
=>2x^2 -x=0
=>x(2x-1)=0
th1 x=0
th2 2x-1=0 => 2x=1 =>x=1/2
b,x^2-1/36=0
<=> x^2-(1/6)^2=0
<=> (x-1/6)(x+1/6)=0
th1 x-1/6=0 =>x=1/6
th2 x+1/6=0 =>x=-1/6
b, (x-2)(x+1)^2 + (x+1)(x-2)^2 = 0
(x-2)(x+1)[(x+1)+(x-2)]=0
(x-2)(x+1)(2x-1)=0
Therefore, three possible answers for x:
(2x-1) = 0, x = 1/2
(x+1) = 0, x = -1
(x-2) = 0, x = 2
X = 2, -1 or 1/2
a. \(-x^2+14x-49=0\Leftrightarrow-\left(x-7\right)^2=0\)
\(\Leftrightarrow x-7=0\Leftrightarrow x=7\)
b. \(\left(x-1\right)^2+2\left(x-1\right)\left(x+2\right)+\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x-1+x+2\right)^2=0\Leftrightarrow2x+1=0\Leftrightarrow x=\dfrac{1}{2}\)
c. \(\left(2x-4\right)\left(3x+1\right)+\left(x-2\right)^2=0\)
\(\Leftrightarrow2\left(x-2\right)\left(3x+1\right)+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)\left(6x+2+x-2\right)=0\Leftrightarrow7x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}7x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
a)-x^2+14x-49=0
-(x^2-2.x.7+7^2)=0
-(x-7)^2=0
x-7=0
x=7
b)(x-1)^2+2(x-1)(x+2)+(x+2)^2=0
(x-1+x+2)^2=0
(2x+1)^2=0
2x+1=0
2x=-1
x=-1/2
c)(2x-4)(3x+1)+(x-2)^2=0
2(x-2)(3x+1)+(x-2)^2=0
(x-2)(2(3x+1)+x-2)=0
(x-2)7x=0
x-2=0 hoặc 7x=0
x=2 hoặc x=0
Bạn xem lại đề câu d nhe
Nếu đúng đề chắc mình ko bit
a) \(5x\left(x+4\right)-x\left(5x+1\right)=0\)
\(\Leftrightarrow x\left[5\left(x+4\right)-5x-1\right]=0\)
\(\Leftrightarrow x\left(5x+20-5x-1\right)=0\Leftrightarrow x=0\)
b) \(3x\left(5-x\right)+4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(4-3x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{4}{3}\end{cases}}\)
c) \(x\left(x-3\right)+4x-12=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)
d) \(x^2-36=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
e) \(x^2+3x+1=2\)
\(\Leftrightarrow x^2+3x+1-2=0\)
\(\Leftrightarrow x^2+3x-1=0\)
\(\Leftrightarrow x^2+3x+\frac{3}{2}-\frac{5}{2}=0\)
\(\Leftrightarrow\left(x+\frac{3}{2}\right)^2-\frac{5}{2}=0\)
\(\Leftrightarrow\left(x+\frac{3}{2}+\frac{\sqrt{5}}{\sqrt{2}}\right)\left(x+\frac{3}{2}-\frac{\sqrt{5}}{\sqrt{2}}\right)=0\)
Còn lại ........... Tự lm nất nha
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
a) x(x-1) - (x+1)(x+2) = 0
x\(^2\)- x -x\(^{^2}\)-2x +x+2=0
-2x+2=0
-2x=0+2
-2x=2
x=-1
Vậy x bằng -1
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
\(\left(4-3x\right)\left(10x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)
\(\left(7-2x\right)\left(4+8x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)
rồi thực hiện đến hết ...
Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>
\(\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)
\(2x^2-7x+3=4x^2+4x-3\)
\(2x^2-7x+3-4x^2-4x+3=0\)
\(-2x^2-11x+6=0\)
\(2x^2+11x-6=0\)
\(2x^2+12x-x-6=0\)
\(2x\left(x+6\right)-\left(x+6\right)=0\)
\(\left(x+6\right)\left(2x-1\right)=0\)
\(x+6=0\Leftrightarrow x=-6\)
\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
\(3x-2x^2=0\)
\(x\left(2x-3\right)=0\)
\(x=0\)
\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Tự lm tiếp nha
Bài 3:
1. \(\left(x-1\right)\left(x+2\right)+5x-5=0\)
\(\Rightarrow\left(x-1\right)\left(x+2\right)+5\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x+2+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
Vậy.......................
2. \(\left(3x+5\right)\left(x-3\right)-6x-10=0\)
\(\Rightarrow\left(3x+5\right)\left(x-3\right)-2\left(3x+5\right)=0\)
\(\Rightarrow\left(3x+5\right)\left(x-3-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x+5=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=5\end{matrix}\right.\)
Vậy........................
3. \(\left(x-2\right)\left(2x+3\right)-7x^2+14x=0\)
\(\Rightarrow\left(x-2\right)\left(2x+3\right)-7x\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(2x+3-7x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\-5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy............................
4, 5 tương tự nhé bn!
bài 3
1 (x-1)(x+2)+5x-5=0
=>(x-1)(x+2)+(5x-5)=o
=>(x-1)(x+2)+5(x-1)=0
=>(x-1)(x+2+5)=0
=>(x-1)(x+7)=0
=>\(\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
vậy x=1 hoặc x=-7
2. (3x+5)(x-3)-6x-10=0
=>(3x+5)(x-3)-(6x+10)=0
=>(3x+5)(x-3)-2(3x+5)=0
=>(3x+5)(x-3-2)=0
=>(3x+5)(x-5)=0
=>\(\left[{}\begin{matrix}3x+5=0\\x-5=0\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=5\end{matrix}\right.\)
a) \(2x^2=x\Rightarrow2x^2-x=0\Rightarrow x\left(2x-1\right)\Rightarrow\left[{}\begin{matrix}x=0\\2x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{2}\end{matrix}\right.\)
b) \(x^2-\frac{1}{36}=0\Rightarrow\left(x+\frac{1}{6}\right)\left(x-\frac{1}{6}\right)=0\Rightarrow\left[{}\begin{matrix}x+\frac{1}{6}=0\\x-\frac{1}{6}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{6}\\x=\frac{1}{6}\end{matrix}\right.\)
c) \(x^2-14x+49=0\Rightarrow\left(x-7\right)^2=0\Rightarrow x-7=0\Rightarrow x=7\)
d) \(x^3-3x^2+3x-1=0\Rightarrow\left(x^3-1\right)-\left(3x^2-3x\right)=0\Rightarrow\left(x-1\right)\left(x^2+x+1\right)-3x\left(x-1\right)=0\Rightarrow\left(x-1\right)\left(x^2+x+1-3x\right)=0\Rightarrow x-1=0\Rightarrow x=-1\)
e) \(x\left(x-2\right)-5x+10=0\Rightarrow x\left(x-2\right)-5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
a) \(2x^2=x\)
\(\Leftrightarrow2x^2-x=0\)
\(\Leftrightarrow2x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
b)\(x^2-\frac{1}{36}=0\)
\(\Leftrightarrow x^2-\left(\frac{1}{6}\right)^2=0\)
\(\Leftrightarrow\left(x-\frac{1}{6}\right)\left(x+\frac{1}{6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{6}=0\\x+\frac{1}{6}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)
c)\(\text{ x²-14x+49=0}\)
\(\Leftrightarrow x^2-2.7.x+7^2=0\)
\(\Leftrightarrow\left(x-7\right)^2=0\)
\(\Leftrightarrow x-7=0\)
\(\Leftrightarrow x=7\)
d)\(x^3-3x^2+3x-1=0\)
\(\Leftrightarrow(x-1)^3=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
\(\text{e) x(x-2) -5x+10=0}\)
\(\Leftrightarrow x\left(x-2\right)-5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)