`2/3 3^x+1 -73^x= -405`
tìm x biết:
a. 2.3^x - 405=3^(x-1)
b. (3/4)^x = 2^8/3^4
c. (5x+1)^2=36/49
d. (x+1)^(x+10)= (x+1)^(x+4)
Tìm x:
( x.1) + (x.2) + (x.3) +....+ (x.9 ) = 405
(1+2+3+4+5+6+7+8+9)xX=405
Số các số hạng:
(9-1):1+1=9 số
Tổng các số hạng:
(1+9)x9:2=45
X=405:45
X=9.
\(\left(x.1\right)+\left(x.2\right)+\left(x.3\right)+...+\left(x.9\right)=405\)
\(x.\left(1+2+3+...+9\right)=405\)
\(x.45=405\)
\(x=405\div45\)
\(x=9\)
tìm x:
3x+2+3x+1+34=405
Tính M=4÷(7×31)+6÷(7×41)+9÷(10×41)+7÷(10×57)
\(3^{x+2}+3^{x+1}+3^4=405\)
\(3^{x+1}.3+3^{x+1}+81=405\)
\(3^{x+1}.\left(3+1\right)=405-81\)
\(3^{x+1}.4=324\)
\(3^{x+1}=81\)
\(3^{x+1}=3^4\)
\(x+1=4\)
\(x=3\)
4 x [ X -12 ] = 2 x X + 164
1\(\frac{1}{4}\)x X - \(\frac{3}{5}\)= \(\frac{4}{5}\)x X + 35%
1/ 3 x 5 + 1/5x9 + ... + 1/401 x 405 + X = 2\(\frac{1}{405}\)
\(4\cdot\left(x-12\right)=2x+164\)
\(\Leftrightarrow4x-48=2x+164\)
\(\Leftrightarrow4x-2x=164+48\)
\(\Leftrightarrow2x=212\Rightarrow x=106\)
Sửa đề : \(\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{401.405}+x=2\frac{1}{405}\)
\(\Leftrightarrow\frac{1}{4}\cdot\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{401}-\frac{1}{405}\right)+x=\frac{811}{405}\)
\(\Leftrightarrow\frac{1}{4}\cdot\left(1-\frac{1}{405}\right)+x=\frac{811}{405}\)
\(\Leftrightarrow\frac{1}{4}\cdot\frac{404}{405}+x=\frac{811}{405}\)
\(\Leftrightarrow\frac{101}{405}+x=\frac{811}{405}\Rightarrow x=\frac{142}{81}\)
4x(X-12)=2xX+164
2x2x(X-12)=2xX+2x82
2x(2X-24)=2x(X+82)
2X-24=X+82
2X=X+82+24
2X=X+106
X=106
Tìm x biết: ( x x 1 ) + ( x x 2 ) + ( x x 3 ) +.....+ ( x x 9 ) = 405
\(\left(x\times1\right)+\left(x\times2\right)+...+\left(x\times9\right)=405\)
\(\Rightarrow x\left(1+2+3+...+9\right)=405\)
\(\Rightarrow x\times45=405\)
\(\Rightarrow x=405:45\)
\(\Rightarrow x=9\)
Xyz sao mình tính được là 405:50 là có dư
( X x 1 ) + ( X x 2 ) + ( X x 3 ) +.....+ ( X x 9 ) = 405
X x ( 1 + 2 +3 + .....+ 9 ) = 405
X x 45 = 405
X = 405: 45
X = 9
2/3.3^x+1-7.3^x=-405
4^x+4^x+3=4160
Tìm x biết:\(\dfrac{2}{3}.3^{x+1}-7.3^x=-405\)
\(\dfrac{2}{3}.3^{x+1}-7.3^x=-405\)
\(\Rightarrow\dfrac{2}{3}.3^x.3^1-7.3^x=-405\)
\(\Rightarrow2.3^x-7.3^x=-405\)
\(\Rightarrow3^x.\left(2-7\right)=-405\)
\(\Rightarrow3^x.-5=-405\)
\(\Rightarrow3^x=\dfrac{-405}{-5}\)
\(\Rightarrow3^x=81\)
Vì \(3^4=81\) nên \(x=4\)
\(\dfrac{2}{3}.3^{x+1}-7.3^x=-405\)
\(\Leftrightarrow\dfrac{2}{3}.3.3^x-7.3^x=-405\)
\(\Leftrightarrow3^x\left(2-7\right)=-405\)
\(\Leftrightarrow3^x.\left(-5\right)=-405\)
\(\Leftrightarrow3^x=81\)
\(\Leftrightarrow3^x=3^4\)
\(\Leftrightarrow x=4\)
Vậy..
Mọi người giúp mình bài tìm x
1) (x- 2/9)^3 = (8/27)^2
2) 2. 3^x - 405 = 3^x- 1
3) 4/3 : 0,8 = 0,75x : (-1,5)
Tìm x
X-1/5 = 2x+1/3
X+3/2x-5 = 2/3
2/3 . 3x+1 - 7 . 3x = -405
2x+3 . 5 + 2x+2 . 3 + 2x+1 . 24 = 200
x- 1/5=2x+1/3
x-2x=1/5+1/3
-x= 8/15
=> x= -8/15
bn ơi đề câu 2 bn viết x+3/2 rồi nhân hay x đó
b: \(\dfrac{x+3}{2x-5}=\dfrac{2}{3}\)
=>4x-10=3x+9
=>x=19
c: \(\Leftrightarrow3^x\cdot\left(\dfrac{2}{3}\cdot3-7\right)=-405\)
=>3^x=81
=>x=4
d: \(\Leftrightarrow2^x\cdot40+2^x\cdot12+2^x\cdot48=200\)
=>2^x=2
=>x=1
Tìm X biết :
$\frac{2}{3}.3^{x+1}-7.3^x=-405$