Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)
=> \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)
=> \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\)
Vậy \(x\in\left\{\frac{9}{20}\right\}\)
\(b,x+\frac{1}{4}=\frac{4}{3}\)
=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)
Vậy \(x\in\left\{\frac{13}{12}\right\}\)
\(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
=> \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\)
=> \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\)
Vậy \(x\in\left\{\frac{25}{42}\right\}\)
\(d,\left|x+5\right|-6=9\)
=> \(\left|x+5\right|=9+6=15\)
=> \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\)
Vậy \(x\in\left\{10;-20\right\}\)
\(e,\left|x-\frac{4}{5}\right|=\frac{3}{4}\)
=> \(\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\)
\(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{6}\)
=> \(\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\)
\(g,x^2=16\)
=> \(\left|x\right|=\sqrt{16}=4\)
=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
vậy \(x\in\left\{4;-4\right\}\)
\(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)
=> \(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)
Vậy \(x\in\left\{\frac{5}{6}\right\}\)
\(i,3^3.x=3^6\)
\(x=3^6:3^3=3^3=27\)
Vậy \(x\in\left\{27\right\}\)
\(J,\frac{1,35}{0,2}=\frac{1,25}{x}\)
=> \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\)
Vậy \(x\in\left\{\frac{5}{27}\right\}\)
\(k,1\frac{2}{3}:x=6:0,3\)
=> \(\frac{5}{3}:x=20\)
=> \(x=\frac{5}{3}:20=\frac{1}{12}\)
Vậy \(x\in\left\{\frac{1}{12}\right\}\)
a,\(\frac{1}{x-1}+\frac{-2}{3}.\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x}\)
\(\Rightarrow\frac{1}{x-1}+\frac{-2}{3}.\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x};Đkxđ:x\ne1\)
\(\Rightarrow\frac{1}{x-1}+\frac{-2}{3}\left(\frac{-9}{20}\right)=\frac{5}{2-2x}\)
\(\Rightarrow\frac{1}{x-1}+\frac{3}{10}=\frac{5}{2-2x}\)
\(\Rightarrow\frac{1}{x-1}-\frac{5}{2-2x}=\frac{-3}{10}\)
\(\Rightarrow\frac{1}{x-1}-\frac{5}{-2\left(x-1\right)}=\frac{-3}{10}\)
\(\Rightarrow\frac{1}{x-1}+\frac{5}{2\left(x-1\right)}=\frac{3}{10}\)
\(\Rightarrow\frac{7}{2\left(x-1\right)}=\frac{-3}{10}\)
\(\Rightarrow70=-6\left(x-1\right)\)
\(\Rightarrow6x=6-70\)
\(\Rightarrow6x=-64\)
\(\Rightarrow x=\frac{-32}{3}x\ne1\)
bài 1 :\(\frac{3}{8}-\frac{1}{5}+\frac{3}{40}=\frac{1}{4}\)
\(\frac{9}{7}\cdot\left(\frac{3}{7}-\frac{1}{2}\right)=-\frac{9}{98}\)
\(-\frac{3}{7}\cdot\frac{5}{9}+\frac{4}{9}\cdot-\frac{3}{7}\cdot\frac{3}{7}=\left(\frac{4}{9}+\frac{5}{9}-1\right)\cdot-\frac{3}{7}=-1\cdot-\frac{3}{7}=\frac{3}{7}\)
bài 2: \(x+\frac{2}{5}=\frac{9}{10}\)
\(x=\frac{9}{10}-\frac{2}{5}\)
\(x=\frac{1}{2}\)
x = \(\frac{9}{10}\)- \(\frac{2}{5}\)
x =\(\frac{9}{10}\) - \(\frac{4}{10}\)
x = \(\frac{5}{10}\) = \(\frac{1}{2}\)
Vậy x = \(\frac{1}{2}\)
ai thấy tớ đúng thì ủng hộ nha
tui đang âm
e) \(\left(x-3\right)\left(x^2+1\right)=0\)
\(\Rightarrow\left(x-3\right)=0\) ( \(x^2+1>0\forall x\))
\(\Rightarrow x=3\)
đ) \(4.8^2=2^x\)
\(2^2.\left(2^3\right)^2=2^x\)
\(2^2.2^6=2^x\)
\(2^8=2^x\)
\(\Rightarrow x=8\)
d) \(\left|x+3\right|=8\)
\(\Rightarrow\orbr{\begin{cases}x+3=8\\x+3=-8\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-11\end{cases}}\)
mấy câu trên dễ rồi tự làm em nhé
Giải:
4.Theo đề bài ta có:
\(A=7.a+4 \)
\(=17.b+3 \)
\(=23.c+11 (a,b,c ∈ N)\)
Nếu ta thêm 150 vào số đã cho thì ta lần lượt có:
\(A+150=7.a+4+150=7.a+7.22=7.(a+22)\)
\(=17.b+3+150=17.b+17.9=17.(b+9)\)
\(=23.c+11+150=23.c+23.7=23.(c+7) \)
\(\Rightarrow A+150⋮7;17;23\).Nhưng 7, 17 và 23 là ba số đôi một nguyên tố cùng nhau, suy ra \(A+150⋮7.17.13=2737\)
Vậy \(A+150=2737k\left(k=1;2;3;4;...\right)\)
Suy ra: \(A=2737k-150=2737k-2737+2587=2737(k-1)+2587=2737k+2587\)
Do \(2587<2737\)
\(\Rightarrow A\div2737\) dư \(2587\)
Bài 1
a) \(\frac{5}{6}=\frac{x-1}{x}\)
<=> 5x=6x-6
<=> 5x-6x=-6
<=> -11x=-6
<=> \(x=\frac{6}{11}\)
b)c)d) nhân chéo làm tương tự
`x : 2^3 = 2^5`
`=> x = 2^5 . 2^3`
`=> x =` \(2^{5+3}\)
`=> x = 2^8`
Vậy `x = 2^8`
`5^9 : x = 5^7`
`=> x = 5^9 : 5^7`
`=>` `x =` \(5^{9-7}\)
`=> x = 5^2`
Vậy `x = 5^2`
`x^2 = 4`
`=> x^2 = 2^2`
`=> x= 2` hoặc `x = -2`
Vậy ` x= 2` hoặc `x = -2`
`4^x = 16`
`=> 4^x = 4^2`
`=> x = 2`
Vậy `x = 2`
a: \(x:2^3=2^5\)
=>\(x=2^5\cdot2^3=2^8=256\)
b: \(5^9:x=5^7\)
=>\(x=5^9:5^7=5^2=25\)
c: \(x^2=4\)
=>\(\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
d: \(4^x=16\)
=>\(4^x=4^2\)
=>x=2