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-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)
![](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) \(\left(x+2\right)^2-\left(3x-7\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=3x-7\\x+2=-3x+7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3x=-2-7\\x+3x=-2+7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=-9\\4x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=\dfrac{5}{4}\end{matrix}\right.\)
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 \(x=\dfrac{1}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(7^{x-5}+121=58\cdot2^3\)
=>\(7^{x-5}+121=58\cdot8\)
=>\(7^{x-5}=343\)
=>x-5=3
=>x=8
b: \(10^{x+25}+168=73\cdot2^4\)
=>\(10^{x+25}+168=73\cdot16\)
=>\(10^{x+25}=1000\)
=>x+25=3
=>x=-22
c: \(5x+59=21\cdot2^2=21\cdot4=84\)
=>5x=84-59=25
=>x=5
d: \(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\)
=>\(2^{x-1}+3=24-16+3\)
=>\(2^{x-1}=8\)
=>x-1=3
=>x=4
7ˣ⁻⁵ + 121 = 58.2³
7ˣ⁻⁵ = 58.8 - 121
7ˣ⁻⁵ = 464 - 121
7ˣ⁻⁵ = 343
7ˣ⁻⁵ = 7³
x - 5 = 3
x = 3 + 5
x = 8
-------
10ˣ⁻²⁵ + 168 = 73.2⁴
10ˣ⁻²⁵ = 73.16 - 168
10ˣ⁻²⁵ = 1000
10ˣ⁻²⁵ = 10³
x - 25 = 3
x = 3 + 25
x = 28
--------
5x + 59 = 21.2²
5x = 21.4 - 59
5x = 84 - 59
5x = 25
x = 25 : 5
x = 5
--------
3 + 2ˣ⁻¹ = 24 - [4² - (2² - 1)]
3 + 2ˣ⁻¹ = 24 - (16 - 3)
2ˣ⁻¹ = 24 - 13 - 3
2ˣ⁻¹ = 8
2ˣ⁻¹ = 2³
2ˣ⁻¹ = 3
x - 1 = 3
x = 3 + 1
x = 4
![](https://rs.olm.vn/images/avt/0.png?1311)
a: -x+5/9=-1/3
nên x=5/9+1/3=5/9+3/9=8/9
b: \(\Leftrightarrow\dfrac{1}{5}+\dfrac{1}{10}+\dfrac{11}{15}< x< \dfrac{1}{2}+\dfrac{13}{12}+\dfrac{1}{3}\)
=>31/30<x<23/12
mà x là số nguyên
nên \(x\in\varnothing\)
c: x+5/3=1/81
nên x=1/81-135/81=-134/81
![](https://rs.olm.vn/images/avt/0.png?1311)
2x - (x + 5) = 300
=> 2x - x - 5 = 300
=> x - 5 = 300
=> x = 305
5x + (x - 40) = 8
=> 5x + x - 40 = 8
=> 6x - 40 = 8
=> 6x = 48
=> x = 8
(x+3).(x-2).2x = 0
<=> x + 3 = 0 => x = -3
<=> x - 2 = 0 => x = 2
<=> 2x = 0 => x = 0
5x - (2x +128) = 314 + 121
=> 5x - 2x - 128 = 435
=> 3x - 128 = 435
=> 3x = 563
=> x = 563/3
x^2 + x = 0
=> x^2 = -x
=> x = 0 hoặc x = -1.
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có
(121-12).(121-22).....(121-112)
= (112-12).(112-22).....(112-112)
= (121-12).(112-22).....0
= 0
( 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_