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2x + 2x + 2 = 160
2x + 2x. 4 = 160
2x(1 + 4) = 160
2x.5 = 160
2x = 160 : 5
2x = 32
2x = 25
=> x = 5
\(2^x+2^{x+2}=160\)
\(2^x+2^x.4=160\)
\(2^x.5=160\)
\(2^x=32=>x=5\)
\(\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+...+\dfrac{1}{\left(x+2\right)\left(x+5\right)}=\dfrac{3}{20}\)
\(\Rightarrow\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{\left(x+2\right)\left(x+5\right)}=\dfrac{3}{20}\)
\(\Rightarrow\dfrac{1}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{\left(x+2\right)\left(x+5\right)}\right)=\dfrac{3}{20}\)
\(\Rightarrow\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+..+\dfrac{3}{\left(x+2\right)\left(x+5\right)}=\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{x+2}-\dfrac{1}{x+5}=\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+5}=\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{x+5}=\dfrac{1}{2}-\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{x+5}=\dfrac{1}{20}\)
\(\Rightarrow x+5=20\)
\(\Rightarrow x=20-5\)
\(\Rightarrow x=15\)
Câu 1 : \(\frac{x+2}{18}+\frac{x+2}{19}+\frac{x+2}{20}=\frac{x+2}{21}+\frac{x+2}{22}\)
=> \(\frac{x+2}{18}+\frac{x+2}{19}+\frac{x+2}{20}-\frac{x+2}{21}-\frac{x+2}{22}=0\)
=> x+2 . ( \(\frac{1}{18}+\frac{1}{19}+\frac{1}{20}-\frac{1}{21}-\frac{1}{22}\)) = 0
Vì \(\frac{1}{18}+\frac{1}{19}_{ }+\frac{1}{20}-\frac{1}{21}-\frac{1}{22}\ne0\)nên x+2=0
=> x= 0 - 2 = -2
Vậy x = -2
a)
Đặt x/2=x/5=k(k thuộc N*) suy ra x=2k và y=5k (1)
Thay (1) vao xy=40 ta được 2k5k=40
10k2=40
k2=4
k= 2 hoặc -2
+) Nếu k=2 thì x=2.2=4 và y=2.5=10
+) Nếu k=-2 thì x=-2.2=-4 và y=-2.5=-10
VẬY (x,y) thuộc {(4,10);(-4,-10)}
x=-0,3y
nên y=-10/3x
Ta có: y=1/2z
nên \(\dfrac{-10}{3}x=\dfrac{1}{2}z\)
\(\Leftrightarrow x=z\cdot\dfrac{1}{2}:\dfrac{-10}{3}=z\cdot\dfrac{1}{2}\cdot\dfrac{-3}{10}=z\cdot\dfrac{-3}{20}\)
a: Khi z=3 thì x=-9/20
Khi z=-2/3 thì \(x=\dfrac{2}{3}\cdot\dfrac{3}{20}=\dfrac{1}{10}\)
Khi z=1/4 thì \(x=\dfrac{-3}{20}\cdot\dfrac{1}{4}=\dfrac{-3}{80}\)
Giải:
Ta có: \(3\left(x-1\right)=2\left(y-2\right)\Rightarrow\frac{x-1}{2}=\frac{y-2}{3}\)
\(4\left(y-2\right)=3\left(z-3\right)\Rightarrow\frac{y-2}{3}=\frac{z-3}{4}\)
\(\Rightarrow\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{2x-2+3y-6}{4+9}=\frac{\left(2x+3y\right)-\left(2+6\right)}{13}=\frac{50-8}{13}=\frac{42}{13}\)
+) \(\frac{x-1}{2}=\frac{42}{13}\Rightarrow x-1=\frac{84}{13}\Rightarrow x=\frac{97}{13}\)
+) \(\frac{y-2}{3}=\frac{42}{13}\Rightarrow y-2=\frac{126}{13}\Rightarrow y=\frac{152}{13}\)
+) \(\frac{z-3}{4}=\frac{42}{13}\Rightarrow z-3=\frac{168}{13}\Rightarrow z=\frac{207}{13}\)
Vậy bộ số \(\left(x;y;z\right)\) là \(\left(\frac{97}{13};\frac{152}{13};\frac{207}{13}\right)\)
Câu a tự làm nhé
b, \(\frac{2x+3}{24}=\frac{3x-1}{32}\)
\(\Leftrightarrow32(2x+3)=24(3x-1)\)
\(\Leftrightarrow64x+96=72x-24\)
\(\Leftrightarrow64x+96-72x=-24\)
\(\Leftrightarrow96-8x=-24\Leftrightarrow x=15\)
2x + 1 - 3.2x - 2 = 40
<=> 2x - 2 + 3 - 3.2x - 2 = 40
<=> 8.2x - 2 - 3.2x - 2 = 40
<=> 5.2x - 2 = 40
<=> 2x - 2 = 8
<=> 2x - 2 = 23
<=> x - 2 = 3
<=> x = 5
Trả lời:
2x + 1 - 3.2x - 2 = 40
<=> 2x . 2 - 3 . 2x : 22 = 40
<=> 2x . 2 - 3 . 2x : 4 = 40
<=> 2x . 2 - 3 . 2x . 1/4 = 40
<=> 2x .( 2 - 3 . 1/4 ) = 40
<=> 2x . 1,25 = 40
<=> 2x = 32
<=> 2x = 25
=> x = 5