Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\dfrac{2x+3}{24}=\dfrac{3x-1}{32}\)
\(\Rightarrow32\left(2x+3\right)=24\left(3x-1\right)\)
\(\Rightarrow64x+96=72x-24\)
\(\Rightarrow8x=120\Rightarrow x=15\)
b) \(\dfrac{13x-2}{2x+5}=\dfrac{76}{17}\)
\(\Rightarrow17\left(13x-2\right)=76\left(2x+5\right)\)
\(\Rightarrow221x-34=152x+380\)
\(\Rightarrow69x=414\Rightarrow x=6\)
a) \(\frac{x-1}{-15}=\frac{-60}{x-1}\)
\(\left(x-1\right)^2=\left(-15\right).\left(-60\right)=900\)
\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^2=300^2\\\left(x-1\right)^2=\left(-300\right)^2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-1=300\\x-1=-300\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=301\\x=-299\end{cases}}\)
b) \(\left|x+\frac{4}{5}\right|+\frac{3}{5}=\frac{2}{5}\)
\(\left|x+\frac{4}{5}\right|=\frac{2}{5}-\frac{3}{5}\)
\(\left|x+\frac{4}{5}\right|=\frac{-1}{5}\)
vì \(\left|x+\frac{4}{5}\right|\ge0\forall x\)mà \(\left|x+\frac{4}{5}\right|=\frac{-1}{5}\)
\(\Rightarrow\)không có giá trị x nào thỏa mãn đề bài trên
c) \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(\Rightarrow x-\frac{1}{2}=\frac{1}{3}\)
\(\Rightarrow x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)
a) \(\Leftrightarrow\left(x-1\right)\left(x-1\right)=\left(-60\right).\left(-15\right)\)
\(\Leftrightarrow\left(x-1\right)^2=900=30^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=30\\x-1=-30\end{cases}\Leftrightarrow\orbr{\begin{cases}x=30+1\\x=-30+1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=31\\x=-29\end{cases}}}\)
Vậy x = 31 hoặc x = - 29
b) \(\left|x+\frac{4}{5}\right|+\frac{3}{5}=\frac{2}{5}\)
\(\Leftrightarrow\left|x+\frac{4}{5}\right|=\frac{2}{5}-\frac{3}{5}\)
\(\Leftrightarrow\left|x+\frac{4}{5}\right|=\frac{-1}{5}\)vô lý không có giá trị tuyệt đối của số nào mà nhận giá trị âm
Vậy ko có giá trị nào của x thỏa mãn
c) \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{3}+\frac{1}{2}\)
\(\Leftrightarrow x=\frac{5}{6}\)
a) Ta có: \(\dfrac{x}{-15}=\dfrac{-60}{x}\)
\(\Leftrightarrow x^2=\left(-15\right)\cdot\left(-60\right)=900\)
hay \(x\in\left\{30;-30\right\}\)
Vậy: \(x\in\left\{30;-30\right\}\)
b) Ta có: \(\left|x\right|+0.573=2\)
\(\Leftrightarrow\left|x\right|=1.427\)
hay \(x\in\left\{1.427;-1.427\right\}\)
Vậy: \(x\in\left\{1.427;-1.427\right\}\)
c) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=-1\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{8}{3};-\dfrac{10}{3}\right\}\)
d) Ta có: \(0.01:2.5=\left(0.75x\right):0.75\)
\(\Leftrightarrow\dfrac{0.75\cdot x}{0.75}=\dfrac{0.01}{2.5}\)
\(\Leftrightarrow x=\dfrac{1}{250}\)
Vậy: \(x=\dfrac{1}{250}\)
Ta có: \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\left(\dfrac{a}{2}\right)^2=\left(\dfrac{b}{3}\right)^2=\left(\dfrac{c}{4}\right)^2\)
\(\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{z^2}{16}\)\(\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{2c^2}{32}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{2c^2}{32}=\dfrac{a^2+b^2-2c^2}{4+9-32}=\dfrac{-76}{-19}=4\)
\(\Rightarrow\dfrac{a^2}{4}=4\Rightarrow a=4\)
\(\dfrac{b^2}{9}=4\Rightarrow b=6\)
\(\dfrac{2c^2}{32}=4\Rightarrow2c^c=128\Rightarrow c=8\)
Vậy \(\left\{{}\begin{matrix}a=4\\b=6\\c=8\end{matrix}\right.\)
a = 4 hoặc a = -4
b = 6 hoặc b = -6
c = 8 hoặc c = -8
a) \(x^2=\left(-15\right).\left(-60\right)=900=>x=\)\(\pm\)\(30\)
b) \(-x^2=\dfrac{-16}{25}=>x^2=\dfrac{16}{25}=>x=\)\(\pm\)\(\dfrac{4}{5}\)
a)\(\dfrac{x}{-15}\)= \(-\dfrac{60}{x}\)
=> x . x = -15 . (-60)
=> \(^{x^2}\) = 900
x = 30
b) \(-\dfrac{2}{x}\) = \(-\dfrac{x}{\dfrac{8}{25}}\)
=> -2 . \(\dfrac{8}{25}\) = x . (-x)
=> \(\dfrac{-16}{25}\) = \(^{x^2}\)
=> x = \(\dfrac{4}{5}\)và \(-\dfrac{4}{5}\)
nhớ tích cho mk vs nha >_<
\(\dfrac{x}{y}=\dfrac{9}{11}\Rightarrow\dfrac{x}{9}=\dfrac{y}{11}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{9}=\dfrac{y}{11}=\dfrac{x+y}{20}=\dfrac{60}{20}=3\\ \Rightarrow\left\{{}\begin{matrix}x=27\\y=33\end{matrix}\right.\)
\(60\%x+\dfrac{2}{3}x=-76\)
\(\dfrac{3}{5}x+\dfrac{2}{3}x=-76\)
\(x\left(\dfrac{3}{5}+\dfrac{2}{3}\right)=-76\)
\(x\dfrac{19}{15}=-76\)
\(x=-76:\dfrac{19}{15}\)
\(x=-60\)
\(60\%x+\dfrac{2}{3}x=-76\)
\(\dfrac{3}{5}x+\dfrac{2}{3}x=-76\)
\(x.\left(\dfrac{3}{5}+\dfrac{2}{3}\right)=-76\)
\(x.\dfrac{19}{15}=-76\)
\(x=-76:\dfrac{19}{15}\)
\(x=-76.\dfrac{15}{19}\)
\(x=-60\)