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ta có \(\frac{x}{12}=\frac{3}{x}\)
\(\Leftrightarrow x^2=12.3=36\)
\(\Leftrightarrow x=6\)hoặc \(x=-6\)
\(3^{2x+4}:3^{x+1}=81\)
\(3^{2x+4-x-1}=3^4\)
\(3^{x+3}=3^4\)
\(\Rightarrow x+3=4\)
\(\Rightarrow x=1\)
1/vì (1,782x-2-1,78x):1,78x=0
nên 1,78x2-2-1,78x=0
=>1,782x-2=1,78x
=>2x-2=x
2x=x+2
=>x=2
2/vì cơ số bằng nhau nên ta có
x-2=1;-1;0
ta có: x-2=1 => x=3
x-2=-1 => x=1
x-2=0 => x=2
3/ta có
(x+2)3=33 =>x+2=3 =>x=1
mik mệt rồi bạn cứ gải tiếp đi
\(3^x+3^{x+2}=90\)
\(3^x+3^x\cdot3^2=90\)
\(3^x\cdot\left(3^2+1\right)=90\)
\(3^x\cdot10=90\)
\(3^x=\frac{90}{10}\)
\(3^x=9\)
\(3^2=9\)
vậy x=2
1) 3^1994+4^1993-3^1992
= 3^1992.(9+3-1)=3^1992.11 chia hết cho 11
=> 3^1994+3^1993-3^1992 chia hết cho 11
a) (x-3)x(4-5x x)=0
=> x-3=0 hoặc 4-5x x=0
=>x=3 hoặc x=0,8
b) x2-2=0
=>x2=2
=>x=\(\sqrt{2}\)
c) x2+\(\sqrt{3}\)=0
=>x2= -\(\sqrt{3}\)
=> Vô nghiệm
d) x2+2x x=0
=> x x(x+2)=0
=> x=0 hoặc x+2=0
=>x=0 hoặc x=-2
e) x2 + 2x x-3=0
=>x2- x+ 3x -3=0
=>(x2-x)+ (3x - 3)=0
=> x(x-1)+ 3(x-1)=0
=>(x-1) x (x+3)=0
=> x-1 =0 hoặc x+3=0
=> x= 1 hoặc x=-3
a) (x-3)x(4-5x x)=0
=> x-3=0 hoặc 4-5x x=0
=>x=3 hoặc x=0,8
b) x2-2=0
=>x2=2
=>x=\(\sqrt{2}\)
c) x2+\(\sqrt{3}\)=0
=>x2= -\(\sqrt{3}\)
=> Vô nghiệm
d) x2+2x x=0
=> x x(x+2)=0
=> x=0 hoặc x+2=0
=>x=0 hoặc x=-2
e) x2 + 2x x-3=0
=>x2- x+ 3x -3=0
=>(x2-x)+ (3x - 3)=0
=> x(x-1)+ 3(x-1)=0
=>(x-1) x (x+3)=0
=> x-1 =0 hoặc x+3=0
=> x= 1 hoặc x=-3
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{2}{3}\right|+\left|x+\frac{3}{4}\right|=4x\)
\(\Rightarrow x=x\)hoặc \(x=-x\)
---Nếu x = x
\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{2}{3}\right|+\left|x+\frac{3}{4}\right|=4x\)
\(x+\frac{1}{2}+x+\frac{2}{3}+x+\frac{3}{4}=4x\)
\(3x+\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}\right)=4x\)
\(\left(\frac{6}{12}+\frac{8}{12}+\frac{9}{12}\right)=4x-3x\)
\(\Rightarrow x=\frac{23}{12}\)
---Nếu x = -x
\(\Rightarrow\left|-x+\frac{1}{2}\right|+\left|-x+\frac{2}{3}\right|+\left|-x+\frac{3}{4}\right|=4x\)
\(-x+\frac{1}{2}+\left(-x\right)+\frac{2}{3}+\left(-x\right)+\frac{3}{4}=4x\)
\(-3x+\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}\right)=4x\)
\(\left(\frac{6}{12}+\frac{8}{12}+\frac{9}{12}\right)=4x+3x\)
\(\Rightarrow7x=\frac{23}{12}\)
\(\Rightarrow x=\frac{23}{12}:7\)
\(\Rightarrow x=\frac{23}{12}.\frac{1}{7}\)
\(\Rightarrow x=\frac{23}{84}\)
\(Vậyx=\frac{23}{12};x=\frac{23}{84}\)
a) \(x+\frac{2}{3}=\frac{4}{5}\)
\(x=\frac{4}{5}-\frac{2}{3}\)
\(x=\frac{2}{15}\)
b) \(x-\frac{2}{7}=\frac{7}{21}\)
\(x=\frac{7}{21}+\frac{2}{7}\)
\(x=\frac{13}{21}\)
c) \(x-\frac{3}{4}=\frac{-8}{11}\)
\(x=\frac{-8}{21}+\frac{3}{4}\)
\(x=\frac{31}{84}\)
d) \(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)
\(\frac{2}{5}+x=\frac{11}{12}-\frac{2}{3}\)
\(\frac{2}{5}+x=\frac{1}{4}\)
\(x=\frac{1}{4}-\frac{2}{5}\)
\(x=\frac{-3}{20}\)
a) \(A=4+4^2+4^3+...+4^{200}\)
\(4A=4^2+4^3+...+4^{201}\)
\(4A-A=3A=4^{201}-4\)
\(A=\frac{4^{201}-4}{3}\)
b) \(B=1+5+5^2+...+5^{2017}\)
\(5B=5+5^2+5^3+...+5^{2018}\)
\(5B-B=4B=5^{2018}-1\)
\(B=\frac{5^{2018}-1}{4}\)
c) \(C=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{500}}\)
\(3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{499}}\)
\(3C-C=2C=1-\frac{1}{3^{500}}=\frac{3^{500}-1}{3^{500}}\)
\(C=\frac{\left(\frac{3^{500}-1}{3^{500}}\right)}{2}\)
T_i_c_k cho mình nha,có j ko hiểu cứ hỏi mình nhé ^^
162:3^2x-4 = 2
3^2x-4 = 162 : 2 = 81 = 3^4
=> 2x-4=4
=> 2x=4+4=8
=> x=8:2=4
3^x.3^3=81
3^x=81:3^3=3 =3^1
=> x=1
k mk nha
162:32x-4=2
32x-4=162:2
32x-4=81
<=> 32x-4=34
=> 2x-4=4
2x=4+4
2x=8
x = 8:2
x = 4
3x.33=81
3x.27=81
3x=81:27
3x=3
<=> x=1