tìm x:
A:23x-18=14x
B:2x+1:16=8
C:2x+4x=20
TÌM X:
A) [124-(20-4x)]:20=12
B)1 phần 3 :(2x-1)= -4 phần 21
C) (1 phần 2.3 + 1 phần 3.4+...+1 phần 8.9 + 1 phần 9.1).x= 1 phần 5
giúp em với ạ biết đáp án là
A)34
B)-3 phần 8
C) 1 phần 2
cần tìm bước giải
A) [124 - (20 - 4x)] = 12 . 20
[124 - (20 - 4x)] = 240
(20 - 4x) = 240 - 124
(20 - 4x) = 116
4x = 116 + 20
4x = 136
x = 136 : 4
x = 34
B) (2x - 1) = 1/3 : -4/21
2x - 1 = -7/4
2x = -7/4 + 1
2x = -3/4
x = -3/4 : 2
x = -3/8
tìm x:
a,32:(3xX-2)=8
b,75:(X-18)=25
c,(15-6xX)x243=729
d,4x(X-12)+9=17
e,20-2x(X+4)=4
a) 32 : (3.x - 2) = 8
3x - 2 = 32 : 8
3x - 2 = 4
3x = 4 + 2
3x = 6
x = 6 : 3
x = 2
b) 75 : (x - 18) = 25
x - 18 = 75 : 25
x - 18 = 3
x = 3 + 18
x = 21
c) (15 - 6.x) . 243 = 729
15 - 6x = 729 : 243
15 - 6x = 3
6x = 15 - 3
6x = 12
x = 12 : 6
x = 2
d) 4.(x - 12) + 9 = 17
4(x - 12) = 17 - 9
4(x - 12) = 8
x - 12 = 8 : 4
x - 12 = 2
x = 2 + 12
x = 14
e) 20 - 2.(x + 4) = 4
2(x + 4) = 20 - 4
2(x + 4) = 16
x + 4 = 16 : 2
x + 4 = 8
x = 8 : 2
x = 4
`32: ( 3xx x -2)=8`
`3xx x-2=32:8`
`3xx x-2=4`
`3 xx x=4+2`
`3xx x=6`
`x=6:3`
`x=2`
__
`75 : (x-18) =25`
`x-18=75:25`
`x-18= 3`
`x=3+18`
`x=21`
__
`(15-6 xx x ) xx 243 =729`
`15-6 xx x = 729 : 243`
`15-6 xx x = 3`
`6 xx x=15-3`
`6 xx x=12`
`x=12:6`
`x=2`
__
`4 xx (x-12)+9=17`
`4 xx (x-12)=17-9`
`4 xx (x-12)= 8`
`x-12=8:4`
`x-12=2`
`x=2+12`
`x=14`
__
`20-2xx(x+4)=4`
`2xx(x+4)=20-4`
`2xx(x+4)=16`
`x+4=16:2`
`x+4=8`
`x=8-4`
`x=4`
\(a,32:\left(3\times X-2\right)=8\\ 3\times X-2=32:8=4\\ 3\times X=4+2=6\\ X=\dfrac{6}{3}=2\\ Vậy:X=2\\ ---\\ b,75:\left(X-18\right)=25\\ X-18=75:25=3\\ X=3+18=21\\ Vậy:X=21\\ ---\\ c,\left(15-6\times X\right)\times243=729\\ 15-6\times X=729:243=3\\ 6\times X=15-3=12\\ X=\dfrac{12}{6}=2\\ Vậy:X=2\\ ---\\ d,4\times\left(X-12\right)+9=17\\ 4\times\left(X-12\right)=17-9=8\\ X-12=8:4=2\\ X=2+12=14\\ Vậy:X=14\\ ---\\ e,20-2\times\left(X+4+4\right)=4\\ 2\times\left(X+4\right)=20-4=16\\ X+4=16:2=8\\ X=8-4=4\\ Vậy:X=4\)
Tìm x:
a)3.(x-2)+2.(x-3)=5
b)(2x-8)2-16=0
c)(2x-1)2-(4x+1).(x-3)=3
a)3(x-2)+2(x-3)=5
=>3x-6+2x-6=5
=>5x=17
=>x=17/5
b)(2x-8)^2=16
TH1:2x-8=4=>x=6
TH2:2x-8=-4=>x=2
\(a,\Leftrightarrow3x-6+2x-6=5\\ \Leftrightarrow5x-12=5\\ \Leftrightarrow x=\dfrac{17}{5}\\ b,\left(2x-8\right)^2-16=0\\ \Leftrightarrow\left(2x-12\right)\left(2x-4\right)=0\\ \Leftrightarrow4\left(x-6\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\\ c,\Leftrightarrow4x^2-4x+1-4x^2+11x+3=3\\ \Leftrightarrow7x=-1\\ \Leftrightarrow x=-\dfrac{1}{7}\)
Tìm x, biết rằng: a) 2x = 16 b) 3x+1 = 9xc) 23x+2 = 4x+5d) 32x-1 = 243
a) 2x = 16 <=>x=8
b) 3x+1 = 9x <=>9x-3x=1
<=>6x=1 <=>x=1/6
c) 23x+2 = 4x+5 <=>23x-4x=5-2
<=>19x=3 <=>x=3/19
d) 32x-1 = 243 <=>32x=244
<=>x=61/8
a/ 2x=16
x=8
b/ 3x+1=9x
3x-9x=-1
-6x=-1
x=1/6
c/ 23x+2=4x
23x-4x=-2
19x=-2
x=-2/19
d/ 32x-1=243
32x=244
x=61/8
Tìm x:
a) 5x(x-2)+(2-x)=0
b) x(2x-5)-10x+25=0
c) \(\dfrac{25}{16}\)-4x2+4x-1=0
d)x4+2x2-8=0
a) \(\text{5x(x-2)+(2-x)=0}\)
\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\text{x(2x-5)-10x+25=0}\)
\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\\ \Rightarrow\left(x-5\right)\left(2x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=2,5\end{matrix}\right.\)
c) \(\dfrac{25}{16}-4x^2+4x-1=0\)
\(\Rightarrow\dfrac{9}{16}-4x^2+4x=0\)
\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)
\(\Rightarrow-4x^2-\dfrac{1}{2}x+\dfrac{9}{2}x+\dfrac{9}{16}=0\)
\(\Rightarrow\left(-4x^2-\dfrac{1}{2}x\right)+\left(\dfrac{9}{2}x+\dfrac{9}{16}\right)=0\)
\(\Rightarrow-\dfrac{1}{2}x\left(8x+1\right)+\dfrac{9}{16}\left(8x+1\right)=0\)
\(\Rightarrow\left(-\dfrac{1}{2}x+\dfrac{9}{16}\right)\left(8x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{2}x+\dfrac{9}{16}=0\\8x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=\dfrac{-1}{8}\end{matrix}\right.\)
a) \(5x\left(x-2\right)+\left(2-x\right)=0\)
\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(x\left(2x-5\right)-10x+25=0\)
\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(2x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{5}{2}\end{matrix}\right.\)
c) \(\dfrac{25}{16}-4x^2+4x-1=0\)
\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)
\(\Rightarrow\left(x-\dfrac{9}{8}\right)\left(x+\dfrac{1}{8}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{9}{8}=0\\x+\dfrac{1}{8}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=-\dfrac{1}{8}\end{matrix}\right.\)
d) \(x^4+2x^2-8=0\)
\(\Rightarrow\left(x^4+2x^2+1\right)-9=0\)
\(\Rightarrow\left(x^2+1\right)^2-3^2=0\)
\(\Rightarrow\left(x^2+1-3\right)\left(x^2+1+3\right)=0\)
\(\Rightarrow\left(x^2-2\right)\left(x^2+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-2=0\\x^2+4=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=2\\x^2=-4\end{matrix}\right.\) \(\Rightarrow x^2=2\) \(\Rightarrow x=\pm\sqrt{2}\)
Tìm x:
a)\(\sqrt{3x-6}\)=3
b)\(\sqrt{5x-16}\)=2
c)Tìm điều kiện xác định của biểu thức: B=\(\dfrac{2x-3}{x^2-4x+3}\)
a) ĐK: x ≥ 2
\(\sqrt{3x-6}=3\)
\(\Leftrightarrow3x-6=9\)
<=> 3x = 15
<=> x = 5
Vậy:....
b) ĐK: 5x - 16 ≥ 0
<=> 5x ≥ 16
<=> x ≥ 16/5
\(\sqrt{5x-16}=2\)
<=> 5x - 16 = 4
<=> 5x = 20
<=> x = 4
c) ĐK: \(x^2-4x+3\ne0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne3\end{matrix}\right.\)
bình phương hai vế ta được:
a)điều kiện của x:x≥2
3x-6=9 <=> x=5(nhận)
b)ĐK: x≥16/5
5x-16=4 <=>x=4(nhận)
c) ta có: \(\dfrac{2x-3}{\left(x-2\right)^2-1}\)= \(\dfrac{2x-3}{\left(x-3\right)\left(x-1\right)}\)
ĐKXĐ: x≠3 ;x≠1
a,\(\sqrt{3x-6}=3\) (với x\(\ge\)2)
=>\(\left(\sqrt{3x-6}\right)^2=3^2\)
<=>\(3x-6=9\)<=>\(3x=9+6\)<=>x=\(\dfrac{15}{3}\)=5(thỏa mãn)
b,\(\sqrt{5x-16}=2\) (với x\(\ge\)16/5)
=>\(\left(\sqrt{5x-16}\right)^2=2^2\)<=>\(5x-16=4< =>5x=20< =>x=4\)(thỏa mãn)
c,B xác định khi \(x^2-4x+3\ne0< =>x^2-2.2.x+2^2-1\ne0\)
\(< =>\left(x-2\right)^2-1\ne0\)
\(< =>\left(x-2+1\right)\left(x-2-1\right)\ne\)0
\(< =>\left(x-1\right)\left(x-3\right)\ne0\)
\(< =>\left[{}\begin{matrix}x-1\ne0\\x-3\ne0\end{matrix}\right.< =>\left[{}\begin{matrix}x\ne1\\x\ne3\end{matrix}\right.\)
Tìm số tự nhiên x:
a) 2x : 4 = 16
b) 4x-3 = 256
c) (2x + 1)3 = 343
d) 10 + 2x = 45 : 43
a) \(2^x:4=16\\ \Rightarrow2^x=64\\ \Rightarrow2^x=2^6\\ \Rightarrow x=6\)
b) \(4^{x-3}=256\\ \Rightarrow4^{x-3}=4^4\\ \Rightarrow x-3=4\\ \Rightarrow x=7\)
c) \(\left(2x+1\right)^3=343\\ \Rightarrow\left(2x+1\right)^3=7^3\\ \Rightarrow2x+1=7\\ \Rightarrow x=3\)
d) \(10+2x=4^5:4^3\\ \Rightarrow10+2x=16\\ \Rightarrow x=3\)
a,2^x:4=16
2^x=16.4=64
2^x=2^6
=>x=6
b,4^x-3=256
4^x-3=4^4
=>x-3=4
x=4+3=7
c,(2x+1)^3=343
(2x+1)^3=7^3
=>2x+1=7
2x=7-1=6
x=6:2=3
d,10+2x=4^5:4^3
10+2x=4^2=16
2x=16-10=6
x=6:2=3
Tìm số tự nhiên x:
a) 2x : 4 = 16
b) 4x-3 = 256
c) (2x + 1)3 = 343
d) 10 + 2x = 45 : 43
a, 2x : 4 = 16
⇒ 2x : 22 = 24
⇒ x - 2 = 4
⇒ x = 6
b, 4x-3 = 256
⇒ 4x - 3 = 44
⇒ x - 3 = 4
⇒ x = 7
c, (2x + 1)3 = 343
⇒ (2x + 1)3 = 73
⇒ 2x + 1 = 7
⇒ 2x = 6
⇒ x = 3
d, 10 + 2x = 45 : 43
⇒ 10 + 2x = 16
⇒ 2x = 6
⇒ x = 3
Tìm x:
a) 3x ( 12x - 4 ) - 9x( 4x - 3 ) = 30
b) x( 5 - 2x) + 2x( x - 1) = 15
`@` `\text {Ans}`
`\downarrow`
`a)`
`3x ( 12x - 4 ) - 9x( 4x - 3 ) = 30`
`=> 3x (12x-4) - 3*3x (4x - 3) = 30`
`=> 3x [12x - 4 - 3(4x-3)] = 30`
`=> 3x (12x - 4 - 12x + 9) = 30`
`=> 3x (-4+9)=30`
`=> 3x*5=30`
`=> 3x=6`
`=> x=2`
Vậy, `x=2`
`b)`
`x( 5 - 2x) + 2x( x - 1)`
`=> x(5-2x) + 2x^2 - 2x=15`
`=> 5x - 2x^2 + 2x^2 - 2x =15`
`=> 3x = 15`
`=> x=5`
Vậy, `x=5.`
a: =>36x^2-12x-36x^2+27x=30
=>15x=30
=>x=2
b: =>5x-2x^2+2x^2-2x=15
=>3x=15
=>x=5