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câu a) mình chịu (dùng kiến thức lớp 12 chắc làm đc haha)
b) gt ⇒ \(\frac{1}{6}.6^{x+2}-6^x=6^{14}-6^{13}\)
⇒ \(6^{x+1}-6^x=6^{14}-6^{13}\)
⇒ \(6^x\left(6-1\right)=6^{13}\left(6-1\right)\)
⇒ \(x=13\)
c) gt ⇒ \(\frac{1}{2}.2^{x+4}-2^x=2^{13}-2^{10}\)
⇒ \(2^{x+3}-2^x=2^{13}-2^{10}\)
⇒ \(2^x\left(2^3-1\right)=2^{10}\left(2^3-1\right)\)
⇒ \(x=10\)
d) gt ⇒ \(\frac{1}{3}.3^{x+4}-4.3^x=3^{16}-4.3^{13}\)
⇒ \(3^{x+3}-4.3^x=3^{16}-4.3^{13}\)
⇒ \(3^x\left(3^3-4\right)=3^{13}\left(3^3-4\right)\)
⇒ \(x=13\)
x=\(\dfrac{4}{15}\) : \(\dfrac{-2}{3}\)
x=\(\dfrac{-2}{5}\)
a: Ta có: \(x\cdot\dfrac{-2}{3}=\dfrac{4}{15}\)
\(\Leftrightarrow x=\dfrac{4}{15}:\dfrac{-2}{3}=\dfrac{4}{15}\cdot\dfrac{-3}{2}=\dfrac{-2}{5}\)
b: Ta có: \(x\cdot\dfrac{-7}{19}=\dfrac{-13}{24}\)
\(\Leftrightarrow x=\dfrac{13}{24}:\dfrac{7}{19}=\dfrac{247}{168}\)
\(\dfrac{x-1}{50}+\dfrac{x-2}{49}=\dfrac{x-3}{48}+\dfrac{x-4}{47}\)
\(\Rightarrow\dfrac{x-1}{50}-1+\dfrac{x-2}{49}-1=\dfrac{x-3}{48}-1+\dfrac{x-4}{47}-1\)
\(\Rightarrow\dfrac{x-51}{50}+\dfrac{x-51}{49}=\dfrac{x-51}{48}+\dfrac{x-51}{47}\)
\(\Rightarrow\dfrac{x-51}{50}+\dfrac{x-51}{49}-\dfrac{x-51}{48}-\dfrac{x-51}{47}=0\)
\(\Rightarrow\left(x-51\right)\left(\dfrac{1}{50}+\dfrac{1}{49}-\dfrac{1}{48}-\dfrac{1}{47}\right)=0\)
Vì \(\dfrac{1}{50}+\dfrac{1}{49}-\dfrac{1}{48}-\dfrac{1}{47}\ne0\) nên \(x-51=0\Rightarrow x=51\)
\(\dfrac{x+25}{6}+\dfrac{x+20}{11}+\dfrac{x+16}{15}+3=0\)
\(\Rightarrow\dfrac{x+25}{6}+1+\dfrac{x+20}{11}+1+\dfrac{x+16}{15}+1=0\)
\(\Rightarrow\dfrac{x+31}{6}+\dfrac{x+31}{11}+\dfrac{x+31}{15}=0\)
\(\Rightarrow\left(x+31\right)\left(\dfrac{1}{6}+\dfrac{1}{11}+\dfrac{1}{15}\right)=0\)
Vì \(\dfrac{1}{6}+\dfrac{1}{11}+\dfrac{1}{15}\ne0\) nên \(x+31=0\Rightarrow x=-31\)
\(\dfrac{x-15}{6}+\dfrac{x-10}{11}=\dfrac{x-3}{18}+\dfrac{x-7}{14}\)
\(\Rightarrow\dfrac{x-15}{6}-1+\dfrac{x-10}{11}-1=\dfrac{x-3}{18}-1+\dfrac{x-7}{14}-1\)
\(\Rightarrow\dfrac{x-21}{6}+\dfrac{x-21}{11}=\dfrac{x-21}{18}+\dfrac{x-21}{14}\)
\(\Rightarrow\dfrac{x-21}{6}+\dfrac{x-21}{11}-\dfrac{x-21}{18}-\dfrac{x-21}{14}=0\)
\(\Rightarrow\left(x-21\right)\left(\dfrac{1}{6}+\dfrac{1}{11}-\dfrac{1}{18}-\dfrac{1}{14}\right)=0\)
Vì \(\dfrac{1}{6}+\dfrac{1}{11}-\dfrac{1}{18}-\dfrac{1}{14}\ne0\) nên \(x-21=0\Rightarrow x=21\)
a) Vì \(\dfrac{x}{y} = \dfrac{5}{3} \Rightarrow \dfrac{x}{5} = \dfrac{y}{3}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\begin{array}{l}\dfrac{x}{5} = \dfrac{y}{3} = \dfrac{{x + y}}{{5 + 3}} = \dfrac{{16}}{8} = 2\\ \Rightarrow x = 2.5 = 10\\y = 2.3 = 6\end{array}\)
Vậy x=10, y=6
b) Vì \(\dfrac{x}{y} = \dfrac{9}{4} \Rightarrow \dfrac{x}{9} = \dfrac{y}{4}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\begin{array}{l}\dfrac{x}{9} = \dfrac{y}{4} = \dfrac{{x - y}}{{9 - 4}} = \dfrac{{ - 15}}{5} = - 3\\ \Rightarrow x = ( - 3).9 = - 27\\y = ( - 3).4 = - 12\end{array}\)
Vậy x = -27, y = -12.
Bài 2:
x=13 nên x+1=14
\(f\left(x\right)=x^{14}-x^{13}\left(x+1\right)+x^{12}\left(x+1\right)-...+x^2\left(x+1\right)-x\left(x+1\right)+14\)
\(=x^{14}-x^{14}-x^{13}+x^{13}-...+x^3+x^2-x^2-x+14\)
=14-x=1
x=13 nên x+1=14
f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14
=x14−x14−x13+x13−...+x3+x2−x2−x+14=x14−x14−x13+x13−...+x3+x2−x2−x+14
=14-x=1
a/ \(\frac{1}{3}.3^x+3^{x+2}=3^{16}+3^{13}\)
\(\Leftrightarrow3^{x-1}+3^{x+2}=3^{13}+3^{16}\)
\(\Leftrightarrow3^{x-1}\left(1+3^3\right)=3^{13}\left(1+3^3\right)\)
\(\Leftrightarrow3^{x-1}=3^{13}\Rightarrow x-1=13\Rightarrow x=14\)
b/ \(\frac{1}{6}6^x+6^{x+2}=6^{15}+6^{18}\)
\(\Leftrightarrow6^{x-1}+6^{x+2}=6^{15}+6^{18}\)
\(\Leftrightarrow6^{x-1}\left(1+6^3\right)=6^{15}\left(1+6^3\right)\)
\(\Rightarrow x=16\)
c/ \(\frac{1}{2}2^{x+3}-2^x=2^{22}-2^{20}\)
\(\Leftrightarrow2^x\left(2^2-1\right)=2^{20}\left(2^2-1\right)\)
\(\Rightarrow x=20\)
7) 5x=4y ⇒\(\dfrac{x}{4}=\dfrac{y}{5}\)
Nhân cả hai vế với \(\dfrac{x}{4}\), ta có: \(\left(\dfrac{x}{4}\right)^2=\dfrac{x}{4}.\dfrac{y}{5}=\dfrac{xy}{20}=\dfrac{20}{20}=1\)
\(\left(\dfrac{x}{4}\right)^2=1\Rightarrow\left[{}\begin{matrix}\dfrac{x}{4}=1\\\dfrac{x}{4}=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}y=5\\y=-5\end{matrix}\right.\)
4) áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{0,5}=\dfrac{y}{0,3}=\dfrac{z}{0,2}=\dfrac{z-y+x}{0,2-0,3+0,5}=\dfrac{1}{\dfrac{2}{5}}=\dfrac{5}{2}\)
\(\dfrac{x}{0,5}=\dfrac{5}{2}\Rightarrow x=\dfrac{5}{4}\)
\(\dfrac{y}{0,3}=\dfrac{5}{2}\Rightarrow y=\dfrac{3}{4}\)
\(\dfrac{z}{0,2}=\dfrac{5}{2}\Rightarrow z=\dfrac{1}{2}\)
6) áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x+11}{13}=\dfrac{y+12}{14}=\dfrac{z+13}{15}=\dfrac{x+11+y+12+z+13}{13+14+15}=\dfrac{42}{42}=1\)
\(\dfrac{x+11}{13}=1\Rightarrow x=2\)
\(\dfrac{y+12}{13}=1\Rightarrow y=1\)
\(\dfrac{z+13}{15}=1\Rightarrow z=2\)
7) \(5x=4y\Rightarrow\dfrac{x}{4}=\dfrac{y}{5}=k\)
\(\Rightarrow x=4k,y=5k\)
\(x.y=20\\ \Rightarrow4k.5k=20\\ \Rightarrow20k^2=20\\ \Rightarrow k^2=1\\ \Rightarrow\left[{}\begin{matrix}k=-1\\k=1\end{matrix}\right.\)
\(x=4k\Rightarrow\left[{}\begin{matrix}x=-4\\x=4\end{matrix}\right.\)
\(y=5k\Rightarrow\left[{}\begin{matrix}y=-5\\y=5\end{matrix}\right.\)
Vậy \(\left(x,y\right)=\left\{\left(-4;-5\right);\left(4;5\right)\right\}\)
1:
a: x/15=-2/6
=>x/15=-1/3
=>x=-5
b: 3/x=1,8/2
=>3/x=9/10
=>x=3*10/9=30/9=10/3
c: (x-3)/(x+2)=2/7
=>2x+4=7x-21
=>-5x=-25
=>x=5
d: (x+1)/3=(x-6)/8
=>8x+8=3x-18
=>5x=-26
=>x=-26/5
e: (2-x)/5=(x+4)/3
=>3(2-x)=5(x+4)
=>5x+20=6-3x
=>8x=-14
=>x=-7/4
g: (2x+1)/(-3)=(1-x)/2
=>2(2x+1)=3(x-1)
=>4x+2=3x-3
=>x=-5
\(\dfrac{x-3}{13}+\dfrac{x-3}{14}=\dfrac{x-3}{15}+\dfrac{x-3}{16}\)
\(\Leftrightarrow\dfrac{1680.\left(x-3\right)+1560.\left(x-3\right)-1456.\left(x-3\right)-1365.\left(x-3\right)}{21840}=0\)
\(\Leftrightarrow\left(x-3\right).\left(1680+1560-1456-1365\right)=0\)
\(\Leftrightarrow\left(x-3\right).419=0\)
\(\Leftrightarrow419x=1257\)
\(\Leftrightarrow x=3\)
Lời giải:
\(\frac{x-3}{13}+\frac{x-3}{14}=\frac{x-3}{15}+\frac{x-3}{16}\)
\((x-3)\left(\frac{1}{13}+\frac{1}{14}\right)=(x-3)\left(\frac{1}{15}+\frac{1}{16}\right)\)
\((x-3)\left[\left(\frac{1}{13}+\frac{1}{14}\right)-\left(\frac{1}{15}+\frac{1}{16}\right)\right]=0\)
Ta thấy:
\(\frac{1}{13}>\frac{1}{15}; \frac{1}{14}>\frac{1}{16}\Rightarrow \frac{1}{13}+\frac{1}{14}> \frac{1}{15}+\frac{1}{16}\)
Do đó biểu thức trong ngoặc vuông lớn hơn $0$ hay khác $0$
$\Rightarrow x-3=0$
$\Leftrightarrow x=3$