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a) \(2\left|x\right|-5=3\)
=> \(2\left|x\right|=3+5\)
=> \(2\left|x\right|=8\)
=> \(\left|x\right|=4\)
=> x = 4 hoặc x = -4
b) \(x-5=\left(-14\right)+2^3\)
=> \(x-5=\left(-14\right)+8\)
=> \(x-5=-6\)
=> \(x=-6+5=-1\)
c) \(10+2x=4^5:4^3\)
=> \(10+2x=4^{5-3}\)
=> \(10+2x=4^2\)
=> \(2x=4^2-10=16-10=6\)
=> \(2x=6\)
=> \(x=3\)
d) (x + 7) - 13 = 4
=> x + 7 = 17
=> x = 17 - 7 = 10
e) \(2x-10=2^4:2^2\)
=> \(2x-10=2^2\)
=> \(2x-10=4\)
=> \(2x=14\)
=> \(x=7\)
`#3107`
b)
`2.3^x = 162`
`\Rightarrow 3^x = 162 \div 2`
`\Rightarrow 3^x = 81`
`\Rightarrow 3^x = 3^4`
`\Rightarrow x = 4`
Vậy, `x = 4`
c)
`(2x - 15)^5 = (2 - 15)^3`
\(\Rightarrow \)`(2x - 15)^5 - (2x - 15)^3 = 0`
\(\Rightarrow \)`(2x - 15)^3 . [ (2x - 15)^2 - 1] = 0`
\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=15\\\left(2x-15\right)^2=\left(\pm1\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x-15=1\\2x-15=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x=16\\2x=-14\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=-7\end{matrix}\right.\)
Vậy, `x \in`\(\left\{-7;8;\dfrac{15}{2}\right\}.\)
`d)`
\(3^{x+2}-5.3^x=?\) Bạn ghi tiếp đề nhé!
`e)`
\(7\cdot4^{x-1}+4^{x-1}=23?\)
\(4^{x-1}\cdot\left(7+1\right)=23\\ \Rightarrow4^{x-1}\cdot8=23\\ \Rightarrow4^{x-1}=\dfrac{23}{8}\)
Bạn xem lại đề!
`f)`
\(2\cdot2^{2x}+4^3\cdot4^x=1056\)
\(\Rightarrow2\cdot2^{2x}+\left(2^2\right)^3\cdot\left(2^2\right)^x=1056\\ \Rightarrow2\cdot2^{2x}+2^6\cdot2^{2x}=1056\\ \Rightarrow2^{2x}\cdot\left(2+2^6\right)=1056\\ \Rightarrow2^{2x}\cdot66=1056\\ \Rightarrow2^{2x}=1056\div66\\ \Rightarrow2^{2x}=16\\ \Rightarrow2^{2x}=2^4\\ \Rightarrow2x=4\\ \Rightarrow x=2\)
Vậy, `x = 2`
_____
\(10 -{[(x \div 3+17) \div 10+3.2^4] \div 10}=5\)
\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=10-5\)
\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=5\)
\(\Rightarrow\left(x\div3+17\right)\div10+48=50\)
\(\Rightarrow\left(x\div3+17\right)\div10=2\)
\(\Rightarrow x\div3+17=20\)
\(\Rightarrow x\div3=3\\ \Rightarrow x=9\)
Vậy, `x = 9.`
\(b)\)\(x\div\left(-5\right)+10=-60\)
\(x\div\left(-5\right)=-60-10\)
\(x\div\left(-5\right)=-70\)
\(x=-70\times\left(-5\right)\)
\(x=350\)
\(a)\)\(3\div x-10=-12\)
\(3\div x=-12+10\)
\(3\div x=-2\)
\(x=3\div-2\)
\(x=-\frac{3}{2}=-1,5\)
tìm x
a.3:x-10=-12
3:x = -12+10
3:x = -2
x = 3:(-2)
x = -3/2
b.x:(-5)+10=-60
x:(-5) =-60-10
x:(-5) = -70
x = -70 x (-5)
x = 350
\(a,\Leftrightarrow x^3=\dfrac{20}{3}\Leftrightarrow x=\sqrt[3]{\dfrac{20}{3}}\\ b,\Leftrightarrow x-1=9\Leftrightarrow x=10\\ c,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow2x+1=5\Leftrightarrow x=2\\ e,\Leftrightarrow2x-4=4\Leftrightarrow x=4\)
Câu a) xem lại đề giùm nhé em
b) \(\left(x-1\right)^3=9^3\)
\(x-1=9\)
\(x=10\)
Vậy \(x=10\)
c) \(\left(x-1\right)^2=25\)
\(x-1=5\) hoặc \(x-1=-5\)
* \(x-1=5\)
\(x=6\)
* \(x-1=-5\)
\(x=-4\)
Vậy \(x=-4\); \(x=6\)
d) \(\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(2x+1=5\)
\(2x=4\)
\(x=2\)
Vậy \(x=2\)
e) Sửa đề: \(\left(2x+4\right)^3=64\)
\(\left(2x+4\right)^3=4^3\)
\(2x+4=4\)
\(2x=0\)
\(x=0\)
Vậy \(x=0\)
a) (x+10)(2y-5) = 143
=> (x+10);(2y-5) thuộc Ư(143)={-1,-143,1,143}
\(\orbr{\begin{cases}x+10=-143\\2y-5=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=-153\\y=2\end{cases}}\)
\(\orbr{\begin{cases}x+10=-1\\2y-5=-143\end{cases}}\Rightarrow\orbr{\begin{cases}x=-11\\y=-69\end{cases}}\)
\(\orbr{\begin{cases}x+10=1\\2y-5=143\end{cases}}\Rightarrow\orbr{\begin{cases}x=-9\\y=74\end{cases}}\)
\(\orbr{\begin{cases}x+10=143\\2y-5=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=133\\y=3\end{cases}}\)
Vậy ta có các cặp x,y thõa mãn : (-153,2);(-11,-69);(-9,74);(113,3)
b) x+(x+1)+(x+2)+..+(x+30)=1240
=> (x+x+x+...+x)+(1+2+3+...+30)=1240
=> 31x+465=1240
31x = 1240-465
31x = 775
x = 775 : 31
x= 25
c) 1+2+3+...+x=210
\(\frac{\left(x-1\right)}{1}+1=x\)
=> \(\frac{\left(x+1\right).x}{2}=210\)
(x+1)x = 210:2
(x+1)x = 105
chắc ko có x thõa mãn
d) 2+4+6+...+2x=210
=> 2(1+2+3+...+x)=210
1+2+3+..+x= 210:2 = 105
\(\frac{\left(x-1\right)}{1}+1\) = x
\(\frac{\left(x+1\right).x}{2}=105\)
(x+1)x = 105:2
(x+1)x = 52,5
ko có x thõa mãn đề bài
a, x + 10 và 2y - 5 thuộc Ư(143) = {1;-1;143;-143}
x + 10 | 1 | -1 | 143 | -143 |
2y - 5 | 143 | -143 | 1 | -1 |
x | -9 | -11 | 133 | -153 |
y | 74 | -69 | 3 | 2 |
b, x+(x+1)+(x+2)+........+(x+30) = 1240
=> x+x+1+x+2+...+x+30=1240
=> 31x+(1+2+...+30) = 1240
=> 31x + 465 = 1240
=> 31x = 775
=> x = 25
c, 1+2+...+x=210
=> \(\frac{x\left(x+1\right)}{2}=210\)
=> x(x+1) = 420
Mà 420 = 20.21
=> x = 20
d, 2+4+...+2x = 210
=> 2(1+2+...+x) = 210
=> \(\frac{2x\left(x+1\right)}{2}=210\)
=> x(x + 1) = 210
Mà 210 = 14.15
=> x = 14
e, 1+3+5+...+(2x-1) = 225
=> \(\frac{\left[\left(2x-1\right)+1\right].x}{2}=225\)
=> \(\frac{2x^2}{2}=225\)
=> x2 = \(\left(\pm15\right)^2\)
=> x = 15 hoặc x = -15
a)
\(\frac{x-3}{10}=\frac{4}{x-3}\)
=> ( x - 3 )2 = 4 . 10.
( x - 3 )2 = 40
Mà x - 3 thuộc Z ( vì x thuộc Z ) nên ( x - 3 )2 là số chính phương.
Do 40 không là số chính phương.
=> Ko tìm được x thuộc Z thỏa mãn đề bài.
b)
\(\frac{x+5}{9}=\frac{4}{x+5}\)
=> ( x + 5 )2 = 4 . 9
( x + 5 )2 = 36
=> x + 5 = 6 hoặc x + 5 = -6.
+) x + 5 = 6
x = 1.
+) x + 5 = -6
x = -11.
Vậy x = 1; x = -11.
Lâu rồi mình ko giải, sai thì thôi nhé!
a) \(\left(10-2x\right)^2=25-\left(-11\right)\)'=
\(\Leftrightarrow\left(10-2x\right)^2=36\)
\(\Leftrightarrow\left(10-2x\right)^2=6^2\)
\(\Leftrightarrow\orbr{\begin{cases}10-2x=6\\10-2x=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=10-6\\2x=10-\left(-6\right)\end{cases}\Leftrightarrow}\orbr{\begin{cases}2x=4\\2x=16\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=8\end{cases}}}\)
Vậy \(x\in\left\{2;8\right\}\)
b) \(-2\left(-x+5\right)-3\left(5-x\right)=4\left(2+x\right)\)
\(\Leftrightarrow-2\left(5-x\right)-3\left(5-x\right)=4\left(2+x\right)\)
\(\Leftrightarrow-5\left(5-x\right)=4\left(2+x\right)\)
\(\Leftrightarrow-25+5x=8+4x\)
\(\Leftrightarrow5x-25=8+4x\)
\(\Leftrightarrow5x=8+4x+25\)
\(\Leftrightarrow5x=4x+33\)
\(\Leftrightarrow5x-4x=33\)
\(\Leftrightarrow1x=33\)
\(\Leftrightarrow x=33\)
Vậy \(x=33\)
a) (10 - 2x)2 = 25 - (-11)
(10 - 2x)2 = 36
(10 - 2x)2 = 62
=> 10 - 2x = 6
2x = 10 - 6
2x = 4
x =4:2
x=2
Vậy x = 2
b)-2(-x+5) - 3(5 - x) = 4(2+x)
2x - 10 - 15 +3x = 8 + 4x
2x - 25 + 3x = 8 +4x
2x + 3x - 4x = 8 + 25
5x - 4x = 33
x= 33
Vậy x = 33
\(\text{a)}3:x-10=-12\)
\(3:x\) \(=\left(-12\right)+10=-2\)
\(x\) \(=3:\left(-2\right)=-1,5\)
\(\text{b)}\left(2-x\right):\left(-4\right)-5=11\)
\(\left(2-x\right):\left(-4\right)\) \(=11+5=16\)
\(\left(2-x\right)\) \(=16x\left(-4\right)=-64\)
\(x\) \(=2-\left(-64\right)=66\)
\(\text{c)}x:\left(-5\right)+10=-60\)
\(x:\left(-5\right)\) \(=\left(-60\right)-10=-70\)
\(x\) \(=\left(-70\right)x\left(-5\right)=350\)
\(\left(3-2x\right)^2=16\)
\(\left(3-2x\right)^2=\pm\left(4\right)^2\)
\(\text{Vậy 3-2x=4 hoặc }\) \(3-2x=-4\)
\(2x=3-4=-1\) \(2x=3-\left(-4\right)=7\)
\(x=\left(-1\right):2=-0,5\) \(x=7:2=3,5\)
\(\text{Câu b với c mik lỡ làm ngược lại rồi bạn thông cảm}\)
\(\text{Hok tốt!}\)
\(\text{@Kaito Kid}\)