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a)\(\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4\)
=> 2x + 7 = 4
2x = 4 - 7
2x = -3
x = -3 : 2
x = -1,5
Vậy x = -1,5
Thay x = -1/3 vào biểu thức A,ta có :
\(\left(-\frac{1}{3}\right)^3-5.\left(-\frac{1}{3}\right)^2+10\)
\(=\left(-\frac{1}{27}\right)-5.\frac{1}{9}+10\)
\(=\left(-\frac{1}{27}\right)-\frac{5}{9}+10\)
\(-\frac{16}{27}+10=\frac{286}{27}\)
Vậy ...
a) \(\frac{4}{9}x+\frac{2}{5}-\frac{1}{3}x=\frac{2}{9}-\frac{1}{4}x\)
\(\Leftrightarrow\frac{13}{36}x=-\frac{8}{45}\)
\(\Rightarrow x=-\frac{32}{65}\)
b) \(\left(\frac{2}{3}x-\frac{1}{2}\right).\left(-\frac{2}{3}\right)+\frac{1}{5}=-\frac{3}{4}\)
\(\Leftrightarrow-\frac{4}{9}x+\frac{1}{3}+\frac{1}{5}=-\frac{3}{4}\)
\(\Leftrightarrow\frac{4}{9}x=\frac{77}{60}\)
\(\Rightarrow x=\frac{231}{80}\)
a) \(\frac{4}{9}x+\frac{2}{5}-\frac{1}{3}x=\frac{2}{9}-\frac{1}{4}x\)
=> \(\frac{4}{9}x-\frac{1}{3}x+\frac{2}{5}-\frac{2}{9}+\frac{1}{4}x=0\)
=> \(\left(\frac{4}{9}x-\frac{1}{3}x+\frac{1}{4}x\right)+\left(\frac{2}{5}-\frac{2}{9}\right)=0\)
=> \(\frac{13}{36}x+\frac{8}{45}=0\)
=> \(\frac{13}{36}x=-\frac{8}{45}\)
=> \(x=-\frac{32}{65}\)
b) \(\left(\frac{2}{3}x-\frac{1}{2}\right)\cdot\frac{-2}{3}+\frac{1}{5}=\frac{-3}{4}\)
=> \(\left(\frac{2}{3}x-\frac{1}{2}\right)\cdot\frac{-2}{3}=-\frac{19}{20}\)
=> \(\frac{2}{3}x-\frac{1}{2}=\left(-\frac{19}{20}\right):\left(-\frac{2}{3}\right)=\left(-\frac{19}{20}\right)\cdot\left(-\frac{3}{2}\right)=\frac{57}{40}\)
=> \(\frac{2}{3}x=\frac{57}{40}+\frac{1}{2}=\frac{77}{40}\)
=> \(x=\frac{77}{40}:\frac{2}{3}=\frac{77}{40}\cdot\frac{3}{2}=\frac{231}{80}\)
a) Đặt \(\frac{x}{-2}=\frac{y}{-3}=k\Rightarrow\hept{\begin{cases}x=-2k\\y=-3k\end{cases}}\)
Khi đó 4x - 3y = 9
<=> -8k + 9k = 9
=> k = 9
=> x = -18 ; y = -27
b) Ta có : \(2x=3y\Rightarrow\frac{2x}{6}=\frac{3y}{6}\Rightarrow\frac{x}{2}=\frac{y}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{2}=\frac{y}{3}=\frac{x+y}{2+3}=\frac{10}{5}=2\)
=> x = 4 ; y = 6
c) Đặt \(\frac{x}{3}=\frac{y}{4}=k\Rightarrow\hept{\begin{cases}x=3k\\y=4k\end{cases}}\)
Khi đó (3k)2 + (4k)2 = 100
<=> 9k2 + 16k2 = 100
=> 25k2 = 100
=> k2 = 4
=> k = \(\pm\)2
Khi k = 2 => x = 6 ; y = 8
Khi k = -2 => x = -6 ; y = -8
Vậy các cặp (x;y) thỏa mãn cần tìm là (6;8);(-6;-8)
d) Đặt \(\frac{x}{3}=\frac{y}{4}=k\Rightarrow\hept{\begin{cases}x=3k\\y=4k\end{cases}}\)
Khi đó x3 + y3 = 91
<=> (3k)3 + (4k)3 = 91
=> 27k3 + 64k3 = 91
=> 91k3 = 91
=> k3 = 1
=> k = 1
=> x = 3 ; y = 4
e) Đặt \(\frac{x}{5}=\frac{y}{4}=k\Rightarrow\hept{\begin{cases}x=5k\\y=4k\end{cases}}\)
Khi đó x2y = 100
<=> (5k)2.4k = 100
=> 25k2.4k = 100
=> 100k3 = 100
=> k = 1
=> x = 5 ; y = 4
biết giải bài 2
x/12=y/14=x.y/12.24=98/288=49/144
=> x/12=49/144=> 49/12
=> y/14=49/144=> 343/72
mới lớp 2 thôi
a)\(\frac{1}{2}x+2\frac{1}{2}=3\frac{1}{2}x-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x=-\frac{3}{4}-\frac{5}{2}\)
\(\Leftrightarrow-3x=-\frac{13}{4}\)
\(\Leftrightarrow x=-\frac{13}{4}:\left(-3\right)\)
\(\Leftrightarrow x=\frac{13}{12}\)
\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x=-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{1}{2}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\Leftrightarrow x=\frac{2}{5}\)
\(c,\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\Leftrightarrow x=-\frac{6}{11}\)
a) \(\left|x-\frac{1}{2}\right|+3=2^2\)
\(\Leftrightarrow\left|x-\frac{1}{2}\right|+3=4\)\(\Leftrightarrow\left|x-\frac{1}{2}\right|=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{2}=-1\\x-\frac{1}{2}=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{3}{2}\end{cases}}\)
Vậy \(x=-\frac{1}{2}\)hoặc \(x=\frac{3}{2}\)
b) \(3^x+3^{x+2}=10^2-2.5\)
\(\Leftrightarrow3^x+3^x.3^2=100-10\)
\(\Leftrightarrow3^x+3^x.9=90\)
\(\Leftrightarrow3^x.\left(1+9\right)=90\)
\(\Leftrightarrow3^x.10=90\)
\(\Leftrightarrow3^x=9=3^2\)\(\Leftrightarrow x=2\)
Vậy \(x=2\)
c) \(3-2x^2=\frac{5}{2}\)
\(\Leftrightarrow2x^2=3-\frac{5}{2}\)\(\Leftrightarrow2x^2=\frac{1}{2}\)
\(\Leftrightarrow x^2=\frac{1}{4}\)\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{1}{2}\end{cases}}\)
Vậy \(x=-\frac{1}{2}\)hoặc \(x=\frac{1}{2}\)
theo mik thì câu a chỉ cần bằng 3/2 là được bạn ah