(5x+1)^2=121
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a) \(\left(3x-1\right)^4=\frac{1}{16}\)
\(\Rightarrow\orbr{\begin{cases}3x-1=\frac{1}{2}\\3x-1=-\frac{1}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=\frac{1}{2}+1=\frac{3}{2}\\3x=-\frac{1}{2}+1=\frac{1}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{1}{6}\end{cases}}\)
b) \(\left(5x-2\right)^2=\frac{25}{121}\)
\(\Rightarrow\orbr{\begin{cases}5x-2=\frac{5}{11}\\5x-2=-\frac{5}{11}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}5x=\frac{5}{11}+2=\frac{27}{11}\\5x=-\frac{5}{11}+2=\frac{17}{11}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{27}{55}\\x=\frac{17}{55}\end{cases}}\)
c) \(\left(5x+6\right)^3=\frac{-8}{27}=\left(\frac{-2}{3}\right)^3\)
\(\Rightarrow5x+6=-\frac{2}{3}\)
\(\Rightarrow5x=-\frac{2}{3}-6=-\frac{20}{3}\)
\(\Rightarrow x=-\frac{20}{3}:5=-\frac{4}{3}\)
a) (3x — 1) ^4 = 1/16
=>3x-1=-1/2 hoặc 1/2
- Với 3x-1=-1/2
=>3x=1/2
=>x=1/6
- Với 3x-1=1/2
=>3x=3/2
=>x=1/2
b)(5x — 2 )^2 = 25/121
=>5x-2=-5/11 hoặc 5/11
- Với 5x-2=-5/11
=>5x=17/11
=>x=17/55
- Với 5x-2=5/11
=>5x=27/11
=>x=27/55
c)(5x + 6)^3= -8/27
=>5x+6=-2/3
=>5x=-20/3
=>x=-4/3
![](https://rs.olm.vn/images/avt/0.png?1311)
-3*(x - 7) - 2(3x + 1) = 5x - 121
=> -3x + 21 - 6x - 2 - 5x + 121 = 0
=> -14x + 140 = 0
=> -14x = -140
=> x = 10
Vậy x = 10
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(\left(4x-1\right)^2-x\left(3-x\right)=121\)
\(\Leftrightarrow16x^2-8x+1-3x+x^2=121\)
\(\Leftrightarrow17x^2-11x-120=0\)
\(\Leftrightarrow17x^2-51x+40x-120=0\)
\(\Leftrightarrow\left(17x^2-51x\right)+\left(40x-120\right)=0\)
\(\Leftrightarrow17x\left(x-3\right)+40\left(x-3\right)=0\)
\(\Leftrightarrow\left(17x+40\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}17x+40=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-40}{17}\\x=3\end{matrix}\right.\)
\(5x^2-x=18\)
\(\Leftrightarrow5x^2-x-18=0\)
\(\Leftrightarrow5x^2-10x+9x-18=0\)
\(\Leftrightarrow\left(5x^2-10x\right)+\left(9x-18\right)=0\)
\(\Leftrightarrow5x\left(x-2\right)+9\left(x-2\right)=0\)
\(\Leftrightarrow\left(5x+9\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+9=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-9}{5}\\x=2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : x= (121-7y)/5
Để x nguyên dương thì 121-7y chia hết cho 5 và 0 < y <18 (y nguyên dương)
để 121-7y chia hết cho 5 thì y=3 hoặc y=13
khi y=13 => x=6
ki y=3 => x= 20
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,121-\left(115+x\right)=3x-\left(25-9-5x\right)-8\\ 121-115-x=3x-25+9+5x-8\\ 6-x=8x-24\\ 8x+x=-24-6\\ 9x=-30\\ x=-\dfrac{30}{9}=-\dfrac{10}{3}\\ ----\\ b,2^{x+2}.3^{x+1}.5^x=10800\\ \left(2.3.5\right)^x.2^2.3=10800\\ 30^x.12=10800\\ 30^x=\dfrac{10800}{12}=900=30^2\\ Vậy:x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x+2\right)^2-\left(3x-7\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=3x-7\\x+2=-3x+7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3x=-2-7\\x+3x=-2+7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=-9\\4x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=\dfrac{5}{4}\end{matrix}\right.\)
Mấy câu kia tương tự.
a) \(\left(x+2\right)^2-\left(3x-7\right)^2=0\)
\(\Leftrightarrow\left(x+2-3x+7\right)\left(x+2+3x-7\right)=0\)
\(\Leftrightarrow\left(-2x+9\right)\left(4x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x+9=0\\4x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=-9\\4x=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-9}{-2}=\dfrac{9}{2}\\x=\dfrac{5}{4}\end{matrix}\right.\)
Vậy \(x=\dfrac{9}{2}\) hoặc \(x=\dfrac{5}{4}\)
b) lộn đề à
c) \(25\left(x-3\right)^2-49\left(2x+1\right)^2=0\)
\(\Leftrightarrow5^2\left(x-3\right)^2-7^2\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[5\left(x-3\right)\right]^2-\left[7\left(2x+1\right)\right]^2=0\)
\(\Leftrightarrow\left(5x-15\right)^2-\left(14x+7\right)^2=0\)
\(\Leftrightarrow\left(5x-15-14x-7\right)\left(5x-15+14x+7\right)=0\)
\(\Leftrightarrow\left(-9x-22\right)\left(19x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-9x-22=0\\19x-8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-9x=22\\19x=8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{22}{-9}=\dfrac{-22}{9}\\x=\dfrac{8}{19}\end{matrix}\right.\)
Vậy \(x=\dfrac{-22}{9}\) hoặc \(x=\dfrac{8}{19}\)
d) \(9\left(3x-2\right)^2=121\left(1-4x\right)^2\)
\(\Leftrightarrow9\left(3x-2\right)^2-121\left(1-4x\right)^2=0\)
\(\Leftrightarrow3^2\left(3x-2\right)^2-11^2\left(1-4x\right)^2=0\)
\(\Leftrightarrow\left[3\left(3x-2\right)\right]^2-\left[11\left(1-4x\right)\right]^2=0\)
\(\Leftrightarrow\left(9x-6\right)^2-\left(11-44x\right)^2=0\)
\(\Leftrightarrow\left(9x-6-11+44x\right)\left(9x-6+11-44x\right)=0\)
\(\Leftrightarrow\left(53x-17\right)\left(-35x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}53x-17=0\\-35x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}53x=17\\-35x=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{53}\\x=\dfrac{-5}{-35}=\dfrac{1}{7}\end{matrix}\right.\)
Vậy \(x=\dfrac{17}{53}\) hoặc \(x=\dfrac{1}{7}\)
( 5x + 1 ) 2 = 121
( 5x + 1 ) = 11
5x = 11 - 1
5x = 10
x = 10 : 5
x = 2
(5x+1)2=121
(5x+1)2=112 hoặc (5x+1)2=(-11)2
5x+1=11 5x+1=-11
5x=10 5x=-12
x=2 x=-12/5
_HT_