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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
11 tháng 6 2018
Bài 1:
Đặt biểu thức trên là A
Ta có:\(A=\left(x-2\right)\left(x+1\right)-\left(x+2\right)\left(x-3\right)=x^2-x-2-\left(x^2-x-6\right)\)
\(=x^2-x-2-x^2+x+6=4\)
Vậy biểu thức A không phụ thuộc vào biến x (đpcm)
Bài 2:
a)\(\left(x-5\right)\left(x+2\right)+\left(x+1\right)\left(2-x\right)=15\)
\(\Leftrightarrow x^2-3x-10+x-x^2+2=15\)
\(\Leftrightarrow-2x-8=15\)
\(\Leftrightarrow-2x=23\)\(\Leftrightarrow x=\frac{-23}{2}\)
Vậy...................................................................................
câu b) tương tự câu a) thôi,bạn tự làm đi nhé
11 tháng 3 2020
ĐKXĐ:\(x\ne\pm2;x\ne-3;x\ne0\)
\(P=1+\frac{x-3}{x^2+5x+6}\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left[\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right]\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left(\frac{2}{x-2}-\frac{x}{x^2-4}-\frac{1}{x+2}\right)\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left[\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\cdot\frac{2x+4-x-x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\frac{8\left(x-3\right)}{\left(x+2\right)^2\left(x+3\right)\left(x-2\right)}\)
Đề sai à ??
30 tháng 9 2018
\(x^3-2x^2-x+2=0\)
\(\Rightarrow x^2\left(x-2\right)-\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x^2-1\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)=0\)
Tìm được \(x\in\left\{2;1;-1\right\}\)
\(\left(x^2+x\right)\left(x^2-x+1\right)=0\)
\(\Rightarrow x\left(x+1\right)\left(x^2-x+1\right)=0\)(1)
Mà \(x^2-x+1=x^2-2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\) (2)
Từ (1) và (2) \(\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
KT
2
6 tháng 7 2017
a) \(25x^2-9=0\)
\(\Leftrightarrow\left(5x\right)^2-3^2=0\)
\(\Leftrightarrow\left(5x+3\right)\left(5x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x+3=0\\5x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{3}{5}\\x=\frac{3}{5}\end{cases}}\)
b) \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
\(\Leftrightarrow x^2+8x+16-x^2+1=16\)
\(\Leftrightarrow8x+17=16\)
\(\Leftrightarrow8x=-1\)
\(\Leftrightarrow x=-\frac{1}{8}\)
6 tháng 7 2017
a) ko hiểu đề bài
b) Ta có (x + 4)2 - (x + 1)(x - 1) = 16
<=> x2 + 8x + 16 - (x2 - 1) = 16
<=> x2 + 8x + 16 - x2 + 1 = 16
<=> 8x + 17 = 16
=> 8x = -1
=> x = \(-\frac{1}{8}\)
17 tháng 7 2018
Lần sau đăng thì chia thành nhiều câu hỏi nhé
\(16^2-9.\left(x+1\right)^2=0\)
\(16^2-\text{ }\left[3.\left(x+1\right)\right]^2=0\)
\(\left[16-3.\left(x+1\right)\right].\left[16+3\left(x+1\right)\right]=0\)
\(\left[16-3x-3\right]\left[16+3x+3\right]=0\)
\(\left[13-3x\right].\left[19+3x\right]=0\)
\(\Rightarrow\orbr{\begin{cases}13-3x=0\\19+3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=13\\3x=-19\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{13}{3}\\x=-\frac{19}{3}\end{cases}}}\)
KL:..............................
HN
19 tháng 2 2019
a) \(-ĐKXĐ:x\ne\pm2;1\)
Rút gọn : \(A=\left(\frac{1}{x+2}-\frac{2}{x-2}-\frac{x}{4-x^2}\right):\frac{6\left(x+2\right)}{\left(2-x\right)\left(x+1\right)}\)
\(=\left(\frac{1}{x+2}+\frac{-2}{x-2}+\frac{x}{x^2-4}\right).\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)
\(=\left[\frac{x-2}{\left(x-2\right)\left(x+2\right)}+\frac{\left(-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x}{\left(x-2\right)\left(x+2\right)}\right]\)\(.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)
\(=\left[\frac{x-2-2x-4+x}{\left(x-2\right)\left(x+2\right)}\right].\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)\(=\frac{x+1}{\left(x+2\right)^2}\)
b) \(A>0\Leftrightarrow\frac{x+1}{\left(x+2\right)^2}>0\Leftrightarrow\orbr{\begin{cases}x+1< 0;\left(x+2\right)^2< 0\left(voly\right)\\x+1>0;\left(x+2\right)^2>0\end{cases}}\)
\(\Leftrightarrow x>1;x>-2\Leftrightarrow x>1\)
Vậy với mọi x thỏa mãn x>1 thì A > 0
c) Ta có : \(x^2+3x+2=0\Leftrightarrow x^2+x+2x+2=0\)
\(\Leftrightarrow x\left(x+1\right)+2\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)
Vậy x = -1;-2
3 tháng 4 2021
a, \(B=\left(\frac{9-3x}{x^2+4x-5}-\frac{x+5}{1-x}-\frac{x+1}{x+5}\right):\frac{7x-14}{x^2-1}\)
\(=\left(\frac{9-3x}{\left(x-1\right)\left(x+5\right)}+\frac{\left(x+5\right)^2}{\left(x-1\right)\left(x+5\right)}-\frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+5\right)}\right):\frac{7\left(x-2\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{9-3x+x^2+10x+25-x^2+1}{\left(x-1\right)\left(x+5\right)}.\frac{\left(x-1\right)\left(x+1\right)}{7\left(x-2\right)}\)
\(=\frac{35+7x}{x+5}\frac{x+1}{7\left(x-2\right)}=\frac{7\left(x+5\right)\left(x+1\right)}{7\left(x+5\right)\left(x-2\right)}=\frac{x+1}{x-2}\)
b, Ta có : \(\left(x+5\right)^2-9x-45=0\)
\(\Leftrightarrow x^2+10x+25-9x-45=0\Leftrightarrow x^2+x-20=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)
TH1 : Thay x = 4 vào biểu thức ta được : \(\frac{4+1}{4-2}=\frac{5}{2}\)
TH2 : THay x = 5 vào biểu thức ta được : \(\frac{5+1}{5-2}=\frac{6}{3}=2\)
c, Để B nhận giá trị nguyên khi \(\frac{x+1}{x-2}\inℤ\Rightarrow x-2+3⋮x-2\)
\(\Leftrightarrow3⋮x-2\Rightarrow x-2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
| x - 2 | 1 | -1 | 3 | -3 |
| x | 3 | 1 | 5 | -1 |
3 tháng 4 2021
d, Ta có : \(B=-\frac{3}{4}\Rightarrow\frac{x+1}{x-2}=-\frac{3}{4}\)ĐK : \(x\ne2\)
\(\Rightarrow4x+4=-3x+6\Leftrightarrow7x=2\Leftrightarrow x=\frac{2}{7}\)( tmđk )
e, Ta có B < 0 hay \(\frac{x+1}{x-2}< 0\)
TH1 : \(\hept{\begin{cases}x+1< 0\\x-2>0\end{cases}\Rightarrow\hept{\begin{cases}x< -1\\x>2\end{cases}}}\)( ktm )
TH2 : \(\hept{\begin{cases}x+1>0\\x-2< 0\end{cases}}\Rightarrow\hept{\begin{cases}x>-1\\x< 2\end{cases}\Rightarrow-1< x< 2}\)
7 tháng 7 2018
\(\Rightarrow25\left(x+1\right)^4-26\left(x+1\right)^2+1=0\Leftrightarrow25\left(x+1\right)^4-25\left(x+1\right)^2-\left(\left(x+1\right)^2-1\right)=0\)
\(\Leftrightarrow25\left(x+1\right)^2.\left(\left(x+1\right)^2-1\right)-\left(\left(x+1\right)^2-1\right)=0\)
\(\Leftrightarrow\left(\left(x+1\right)^2-1\right).\left(25\left(x+1\right)^2-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x+1\right)^2-1=0\\25\left(x+1\right)^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0,-2\\x=-\frac{4}{5},-\frac{6}{5}\end{cases}}}\)
7 tháng 7 2018
\(x^2+x-1=0\Leftrightarrow\left(x+\frac{1}{2}\right)^2-\frac{5}{4}=0\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{\sqrt{5}}{2}\\x+\frac{1}{2}=\frac{-\sqrt{5}}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{5}-1}{2}\\x=\frac{-\sqrt{5}-1}{2}\end{cases}}}\)
\(x^3-x^2-x^2+x=0\Leftrightarrow x^3-2x^2+x=0\)
\(\Leftrightarrow x\left(x^2-2x+1\right)=0\Leftrightarrow x\left(x-1\right)^2=0\Leftrightarrow x=0;x=1\)