LM
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
DL
5
MD
9 tháng 6 2017
f) \(4x^2-12x+9=0\)
<=> \(\left(2x-3\right)^2\) = 0
<=> \(2x-3=0\)
<=> \(2x=3\) <=> \(x=\dfrac{3}{2}\)
Vậy ...............
g) \(3x^2+7x+2=0\)
<=> \(\left(3x^2+6x\right)+\left(x+2\right)=0\)
<=> \(3x\left(x+2\right)+\left(x+2\right)=0\)
<=> \(\left(x+2\right)\left(3x+1\right)=0\)
<=> \(\left[{}\begin{matrix}x=-2\\x=\dfrac{-1}{3}\end{matrix}\right.\)
Vậy ........................
h) \(x^2-4x+1=0\)
<=> \(\left(x^2-4x+4\right)-3=0\)
<=> \(\left(x-2\right)^2=3\)
<=> \(\left[{}\begin{matrix}x+2=\sqrt{3}\\x+2=-\sqrt{3}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=\sqrt{3}-2\\x=-\sqrt{3}-2\end{matrix}\right.\)
Vậy .........................
i) \(2x^2-6x+1=0\)
<=> \(2\left(x^2-3x+2,25\right)-3,5=0\)
<=> \(\left(x-1,5\right)^2=1,75\)
<=> \(\left[{}\begin{matrix}x-1,5=\sqrt{1,75}\\x-1,5=-\sqrt{1,75}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=\sqrt{1,75}+1,5\\x=-\sqrt{1,75}+1,5\end{matrix}\right.\)
Vậy ...................
j) \(3x^2+4x-4=0\)
<=> \(\left(3x^2+6x\right)-\left(2x+4\right)=0\)
<=> \(3x\left(x+2\right)-2\left(x+2\right)\) = 0
<=> \(\left(x+2\right)\left(3x-2\right)=0\)
<=> \(\left[{}\begin{matrix}x=-2\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy ....................................
9 tháng 6 2017
f) \(4x^2-12x+9=0\)
\(\Rightarrow\left(2x-3\right)^2=0\)
\(\Rightarrow2x-3=0\)
\(\Rightarrow x=\dfrac{3}{2}\)
Vậy..
g) \(3x^2+7x+2=0\)
\(\Rightarrow3x^2+6x+x+2=0\)
\(\Rightarrow3x\left(x+2\right)+\left(x+2\right)=0\)
\(\Rightarrow\left(x+2\right)\left(3x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\3x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1}{3}\end{matrix}\right.\)
Vậy..
h) \(x^2-4x+1=0\)
\(\Rightarrow x^2-4x+4-3=0\)
\(\Rightarrow\left(x-2\right)^2-3=0\)
\(\Rightarrow\left(x-2-\sqrt{3}\right)\left(x-2+\sqrt{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2-\sqrt{3}=0\\x-2+\sqrt{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2+\sqrt{3}\\x=2-\sqrt{3}\end{matrix}\right.\)
Vậy..
j) \(3x^2+4x-4=0\)
\(\Rightarrow3x^2+6x-2x-4=0\)
\(\Rightarrow3x\left(x+2\right)-2\left(x+2\right)=0\)
\(\Rightarrow\left(x+2\right)\left(3x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy..
DB
18 tháng 6 2020
a, (x-5).(x-1) >0
<=> x-5>0 và x-1>0
<=> x-5>0
<=> x>5
x-1>0
<=> x>1
Vậy x>5
b, (2x-3).(x+1) <0
<=> 2x-3<0 và x+1<0
2x-3<0 <=> 2x<3 <=> x<2/3
x+1<0 <=> x<-1
Vậy x<2/3
c, 2x2 - 3x +1>0
<=> 2x2 - 2x- x +1>0
<=>(x-1). (2x-1) >0
<=> x-1>0 và 2x-1>0
x-1>0 <=> x>1
2x-1>0 <=> 2x>1 <=> x>1/2
Vậy x>1/2
7 tháng 11 2021
e) \(\left(9x^2-49\right)+\left(3x+7\right)\left(7x+3\right)=0\)
\(\Rightarrow\text{[}\left(3x\right)^2-7^2\text{]}+\left(3x+7\right)\left(7x+3\right)=0\)
\(\Rightarrow\left(3x-7\right)\left(3x+7\right)+\left(3x+7\right)\left(7x+3\right)=0\)
\(\Rightarrow\left(3x+7\right)\text{[}\left(3x-7\right)+\left(7x+3\right)\text{]}=0\)
\(\Rightarrow\left(3x+7\right)\left(3x-7+7x+3\right)=0\)
\(\Rightarrow\left(3x+7\right)\left(10x-4\right)=0\)
=> 2 TH
*3x+7=0 *10x-4=0
=>3x=-7 =>10x=4
=>x=-7/3 =>x=4/10=2/5
vậy x=-7/3 hoặc x=2/5
g) \(\left(x-4\right)^2=\left(2x-1\right)^2\)
\(\Rightarrow\left(x-4\right)^2-\left(2x-1\right)^2=0\)
\(\Rightarrow\left(x-4-2x+1\right)\left(x-4+2x-1\right)=0\)
\(\Rightarrow\left(-x-3\right)\left(3x-5\right)=0\)
\(\Rightarrow-\left(x+3\right)\left(3x-5\right)=0\)
=> 2 TH
*-(x+3)=0 *3x-5=0
=>-x=-3 =>3x=5
=x=3 =>x=5/3
h)\(x^2-x^2+x-1=0\)
\(\Rightarrow0+x-1=0\)
\(\Rightarrow x-1=0\)
=>x=0+1
=>x=1
vậy x=1
k, x(x+ 16) - 7x - 42 = 0
=>x^2+16x-7x-42=0
=>x^2+9x-42=0
vì x^2>0
do đó x^2+9x-42>0
nên o có gt nào của x t/m y/cầu đề bài
m)x^2+7x+12=0
=>x^2+3x++4x+12=0
=>x(x+3)+4(x+3)=0
=>(x+4).(x+3)=0
=>2 TH
=> *x+4=0
=>x=-4
vậy x=-4
*x+3=0
=>x=-3
vậy x=-3
n)x^2-7x+12=0
=>x^2-4x-3x+12=0
=>x(x-4)-3(x-4)=0
=>(x-3).(x-4)=0
=>2 TH
*x-3=0=>x=0+3=>x=3
*x-4=0=>x=0+4=>x=4
vậy x=3 hoặc x=4
KA
7 tháng 11 2021
a)(3x−3)(5−21x)+(7x+4)(9x−5)=44⇔15x−63x2−15+63x+63x2−35x+36x−20=44⇔79x−35=44⇔79x=79⇒x=1a)(3x−3)(5−21x)+(7x+4)(9x−5)=44⇔15x−63x2−15+63x+63x2−35x+36x−20=44⇔79x−35=44⇔79x=79⇒x=1
b)(x+1)(x+2)(x+5)−x2(x+8)=27⇔x2+2x+x+2(x+5)−x3−8x2=27⇔x2(x+5)+2x(x+5)+x(x+5)+2(x+5)−x3−8x2=27⇔x3+5x2+2x2+10x+x2+5x+2x+10−x3−8x2=27⇔17x+10=27⇔17x=17⇒x=1
19 tháng 3 2020
c, x+x4=0
=>x(x+3)=0
=>x=0 hoặc x+3=0
=>x=0 hoặc x = -3
ST
12 tháng 2 2020
Mik mới làm có bằng này bạn xem còn căc ý còn lại mik sẽ có làm.
Q
7 tháng 6 2017
n) \(\left|3-x\right|+x^2-x\left(x+4\right)=0\)
\(\Rightarrow\left|3-x\right|+x^2-x^2-4x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3-x-4x=0\left(đk:3-x\ge0\right)\\-\left(3-x\right)-4x=0\left(đk:3-x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(đk:x\le3\right)\\x=-1\left(đk:x>3\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x=\dfrac{3}{5}\)
m) \(\left(x-1\right)^2+\left|x+21\right|-x^2-13=0\)
\(\Rightarrow x^2-2x+1+\left|x+21\right|-x^2-13=0\)
\(\Leftrightarrow-2x-12+\left|x+21\right|=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x-12+x+21=0\left(đk:x+21\ge0\right)\\-2x-12-\left(x+21\right)=0\left(đk:x+21< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\left(đk:x\ge-21\right)\\x=-11\left(đk:x< -21\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=9\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x=9\)
Q
7 tháng 6 2017
e) \(\left|5x\right|=3x-2\)
\(\Rightarrow5\cdot\left|x\right|=3x-2\)
\(\Leftrightarrow5\cdot\left|x\right|-3x=-2\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-3x=-2\left(đk:x\ge0\right)\\5\cdot\left(-x\right)-3x=-2\left(đk:x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(đk:x\ge0\right)\\x=\dfrac{1}{4}\left(đk:x< 0\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x\in\varnothing\)
g) \(\left|-2,5x\right|=x-12\)
\(\Rightarrow2,5\cdot\left|x\right|=x-12\)
\(\Leftrightarrow2x5\cdot\left|x\right|-x=-12\)
\(\Leftrightarrow\left[{}\begin{matrix}2,5x-x=-12\left(đk:x\ge0\right)\\2,5\cdot\left(-x\right)-x=-12\left(đk:x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-8\left(đk:x\ge0\right)\\x=\dfrac{24}{7}\left(đk:x< 0\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x\in\varnothing\)
GH
11 tháng 2 2018
a) \(\dfrac{x+1}{2}+\dfrac{3x-2}{3}=\dfrac{x-7}{12}\)
\(\Leftrightarrow\dfrac{6\left(x+1\right)+4\left(3x-2\right)}{12}=\dfrac{x-7}{12}\)
\(\Leftrightarrow6\left(x+1\right)+4\left(3x-2\right)=x-7\)
\(\Leftrightarrow6x+6+12x-8=x-7\)
\(\Leftrightarrow6x+12x-x=-7-6+8\)
\(\Leftrightarrow17x=-5\)
\(\Leftrightarrow x=\dfrac{-5}{17}\)
Vậy .........................
b) \(\dfrac{2x}{x-3}-\dfrac{5}{x+3}=\dfrac{x^2+21}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\)
\(\Leftrightarrow\dfrac{2x\left(x+3\right)-5\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+21}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow2x\left(x+3\right)-5\left(x-3\right)=x^2+21\)
\(\Leftrightarrow2x^2+6x-5x+15=x^2+21\)
\(\Leftrightarrow2x^2-x^2+x+15-21=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow x^2-2x+3x-6=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(n\right)\\x=-3\left(l\right)\end{matrix}\right.\)
Vậy \(S=\left\{2\right\}\)
d) \(\left(x-4\right)\left(7x-3\right)-x^2+16=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x^2-16\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3-x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(6x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\6x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{6}\end{matrix}\right.\)
Vậy .........................
P/s: các câu còn lại tương tự, bn tự giải nha
f) sai đề đúng không???
g) 3x - 11 = 0
3x = 11
x = \(\frac{11}{3}\)
Vậy ptrình có nghiệm x = \(\frac{11}{3}\)
h) 2x + x + 12 = 0
3x + 12 = 0
3x = -12
x = -4
Vậy ptrình có nghiệm x = -4