Giải phương trình sau
80/x+4 + 80/x-4 = 25/3
K
Khách
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18 tháng 3 2023
\(\Leftrightarrow\dfrac{16}{x+4}+\dfrac{16}{x-4}=\dfrac{5}{3}\)
=>\(\dfrac{16x-64+16x+64}{x^2-16}=\dfrac{5}{3}\)
=>5(x^2-16)=3*32x=96x
=>5x^2-96x-80=0
=>x=20 hoặc x=-4/5
19 tháng 3 2023
nếu là giải PT bằng cách quy đồng:
25x2 + 480 - 400 = 0
làm sao để phan tích ra ạ.
4 tháng 4 2020
\(\frac{80}{x+4}+\frac{80}{x-4}=\frac{25}{3}\left(ĐkXĐ:x\ne\pm4\right)\\ \Leftrightarrow\frac{3.80\left(x-4\right)+3.80\left(x+4\right)-25.\left(x^2-16\right)}{\left(x^2-16\right).3}=0\\ \Leftrightarrow\frac{240x-960+240x+960-25x^2+400}{3.\left(x^2-16\right)}=0\\ \Leftrightarrow\frac{-25x^2+480x+400}{3.\left(x^2-16\right)}=0\\ \Leftrightarrow\frac{-25.\left(x+\frac{4}{5}\right)\left(x-20\right)}{3.\left(x^2-16\right)}=0\\ \Leftrightarrow\left[{}\begin{matrix}x+\frac{4}{5}=0\\x-20=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\frac{4}{5}\left(Nhận\right)\\x=20\left(Nhận\right)\end{matrix}\right.\\ \Rightarrow S=\left\{-\frac{4}{5};20\right\}\)
RK
28 tháng 8 2021
a) \(x^4-x^2+\dfrac{1}{4}-\dfrac{225}{4}=0\\ \left(x^2-\dfrac{1}{2}\right)^2-\dfrac{15}{2}^2=0\\ \left(x+7\right)\left(x-8\right)=0\\ \left[{}\begin{matrix}x=8\\x=-7\end{matrix}\right.\)
Vậy x = 8 hoặc x = -7
28 tháng 8 2021
a: Ta có: \(x^4-x^2-56=0\)
\(\Leftrightarrow x^4-8x^2+7x^2-56=0\)
\(\Leftrightarrow\left(x^2-8\right)\left(x^2+7\right)=0\)
\(\Leftrightarrow x^2-8=0\)
hay \(x\in\left\{2\sqrt{2};-2\sqrt{2}\right\}\)
QN
1
23 tháng 12 2023
ĐKXĐ: x>=-1
Sửa đề: \(6\sqrt{x+1}-\sqrt{25x+25}+8\sqrt{\dfrac{x+1}{4}}=10\)
=>\(6\sqrt{x+1}-5\sqrt{x+1}+8\cdot\dfrac{\sqrt{x+1}}{2}=10\)
=>\(\sqrt{x+1}+4\sqrt{x+1}=10\)
=>\(5\sqrt{x+1}=10\)
=>\(\sqrt{x+1}=2\)
=>x+1=4
=>x=3(nhận)
18 tháng 1 2016
a, <=> (x-5/100) -1 +(x-4/101) -1 +(x-3/102) -1= (x-100/5) -1+(x-101/4) -1 +(x-102/3) -1
<=> (x-105)(1/100 +1/101 +1/102)= (x-105)(1/5+1/4+1/3)
<=> (x-105)(1/100+1/101+1/102-1/5-1/4-1/3)=0
vì 1/100+1/101+1/102-1/5-1/4-1/3 khác 0 <=> x-105=0
<=> x=105
18 tháng 1 2016
b, 29-x/21 +1+27-x/23 +1+25-x/25 +1+23-x/27 +1+21-x/29 +1=0
<=> 50-x/21 +50-x/23 +50-x/25 +50-x/27 +50-x/29=0
<=> (50-x)(1/21 +1/23 +1/25 +1/27 +1/29)=0
vì 1/21+1/23+1/25+1/27+1/29 lớn hơn 0
nên 50-x=0
<=> x=50
22 tháng 4 2022
a.\(x^2-25=8\left(5-x\right)\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)-8\left(5-x\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)+8\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-13\end{matrix}\right.\)
b.\(\dfrac{x-2}{x+2}-\dfrac{2\left(x-11\right)}{x^2-4}=\dfrac{3}{x-2}\) ; \(ĐK:x\ne\pm2\)
\(\Leftrightarrow\dfrac{\left(x-2\right)\left(x-2\right)-2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\left(x-2\right)^2-2\left(x-11\right)=3\left(x+2\right)\)
\(\Leftrightarrow x^2-4x+4-2x+22=3x+6\)
\(\Leftrightarrow x^2-9x+20=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=4\left(tm\right)\end{matrix}\right.\)
25 tháng 2 2022
a,\(\left(x-4-5\right)\left(x-4+5\right)=0\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\Leftrightarrow x=9;x=-1\)
b, \(\left(x-3-x-1\right)\left(x-3+x+1\right)=0\Leftrightarrow2x-2=0\Leftrightarrow x=1\)
c, \(\left(x^2-4\right)\left(2x-3\right)-\left(x^2-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(2x-3-x+1\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-2\right)=0\Leftrightarrow x=-2;x=2\)
d, \(\left(3x-7\right)^2-\left(2x+2\right)^2=0\Leftrightarrow\left(3x-7-2x-2\right)\left(3x-7+2x+2\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(5x-5\right)=0\Leftrightarrow x=1;x=9\)
BN
25 tháng 2 2022
a) Ta có: 4x-20=0
⇔4x=20⇔4x=20
hay x=5
Vậy: S={5}
b) Ta có: 2x+x+12=02x+x+12=0
⇔3x+12=0⇔3x+12=0
⇔3x=−12⇔3x=−12
hay x=-4
\(\frac{80}{x+4}+\frac{80}{x-4}=\frac{25}{3}\left(1\right)\)
ĐKXĐ: \(x\ne\pm4\)
\(\left(1\right)\Leftrightarrow\frac{80\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}+\frac{80\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\frac{25}{3}\)
\(\Leftrightarrow\frac{80x-320+80x+320}{\left(x-4\right)\left(x+4\right)}=\frac{25}{3}\)
\(\Leftrightarrow\frac{160x}{\left(x-4\right)\left(x+4\right)}=\frac{25}{3}\)
\(\Rightarrow480x=25\left(x-4\right)\left(x+4\right)\)
\(\Leftrightarrow480x=25\left(x^2-16\right)\)
\(\Leftrightarrow480x=25x^2-400\)
\(\Leftrightarrow25x^2-480x-400=0\)
\(\Leftrightarrow25x^2-480x+2304-2704=0\)
\(\Leftrightarrow\left(5x-48\right)^2=2704\)
\(\Leftrightarrow\left|5x-48\right|=52\)
+Trường hợp 1: Nếu \(x\ge\frac{48}{5}\)thì \(5x-48\ge0\Rightarrow\left|5x-48\right|=5x-48\)
Ta có phương trình;
\(5x-48=52\)
\(\Leftrightarrow5x=100\)
\(\Leftrightarrow x=20\)(thỏa măn)
+Trường hợp 2: Nếu\(x< \frac{48}{5}\)thì \(5x-48< 0\Rightarrow\left|5x-48\right|=48-5x\)
Ta có phương trình:
\(48-5x=52\)
\(\Leftrightarrow-5x=4\)
\(\Leftrightarrow x=-\frac{4}{5}\)(thỏa mãn)
\(S=\left\{20;-\frac{4}{5}\right\}\)
Ta có: \(\frac{80}{x+4}+\frac{80}{x-4}=\frac{25}{3}\)
\(\Leftrightarrow80.\left[\frac{x-4+x+4}{\left(x+4\right).\left(x-4\right)}\right]=\frac{25}{3}\)
\(\Leftrightarrow\frac{2x}{x^2-16}=\frac{25}{3}:80\)
\(\Leftrightarrow\frac{2x}{x^2-16}=\frac{5}{48}\)
\(\Rightarrow48.2x=5.\left(x^2-16\right)\)
\(\Leftrightarrow96x=5x^2-80\)
\(\Leftrightarrow5x^2-96x-80=0\)
\(\Leftrightarrow\left(5x^2-100x\right)+\left(4x-80\right)=0\)
\(\Leftrightarrow5x.\left(x-20\right)+4.\left(x-20\right)=0\)
\(\Leftrightarrow\left(5x+4\right).\left(x-20\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x+4=0\\x-20=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}5x=-4\\x=20\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{4}{5}\left(TM\right)\\x=20\left(TM\right)\end{cases}}\)
Vậy \(S=\left\{-\frac{4}{5};20\right\}\)