Tìm x ,biết : \(\frac{3}{4}\) + \(\frac{2}{5}\)x = \(\frac{29}{60}\)
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14 tháng 10 2020
a) \(\left|2x+\frac{3}{4}\right|=\frac{1}{2}\)
\(\orbr{\begin{cases}2x+\frac{3}{4}=\frac{1}{2}\\2x+\frac{3}{4}=\frac{-1}{2}\end{cases}}\) => \(\orbr{\begin{cases}2x=\frac{1}{2}-\frac{3}{4}\\2x=\frac{-1}{2}-\frac{3}{4}\end{cases}}\) => \(\orbr{\begin{cases}2x=\frac{-1}{4}\\2x=\frac{-5}{4}\end{cases}}\) => \(\orbr{\begin{cases}x=\frac{-1}{8}\\x=\frac{-5}{8}\end{cases}}\)
Vậy \(x=\left\{\frac{-1}{8},\frac{-5}{8}\right\}\)
b) \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{2\frac{1}{4}}\)= \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{\frac{9}{4}}\)
=> \(3x.\frac{9}{4}=2,7.\frac{1}{4}\)=> \(\frac{27x}{4}=\frac{27}{40}\)
\(27x.40=27.4\)
\(1080.x=108\)
\(x=\frac{1}{10}\)
Vậy \(x=\frac{1}{10}\)
c) \(\left|x-1\right|+4=6\)
\(\left|x-1\right|=6-4\)
\(\left|x-1\right|=2\)
\(\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\)=> \(\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
Vậy \(x=\left[3,-1\right]\)
d) \(\frac{x}{3}=\frac{y}{5}=>\frac{y}{5}=\frac{x}{3}=>\frac{y-x}{5-3}=\frac{24}{2}=12\)
e) \(\left(x^2-3\right)^2=16\)
\(\left(x^2-3\right)^2=4^2\)\(=>x^2-3=4\)
\(x^2=7=>x=\sqrt{7}\)
Vậy \(x=\sqrt{7}\)
f) \(\frac{3}{4}+\frac{2}{5}x=\frac{29}{60}\)
\(\frac{2}{5}x=\frac{29}{60}-\frac{3}{4}\)
\(\frac{2}{5}x=-\frac{4}{15}\)
\(x=-\frac{4}{15}:\frac{2}{5}=-\frac{4}{15}.\frac{5}{2}=-\frac{2}{3}\)
Vậy \(x=-\frac{2}{3}\)
g) \(\left(-\frac{1}{3}\right)^3.x=\frac{1}{81}\)
\(\left(-\frac{1}{27}\right).x=\frac{1}{81}\)
\(x=\left(-\frac{1}{27}\right):\frac{1}{81}=\left(-\frac{1}{27}\right).81=-3\)
Vậy \(x=-3\)
k)\(\frac{3}{4}-\frac{2}{5}x=\frac{29}{60}\)
\(\frac{2}{5}x=\frac{3}{4}-\frac{29}{60}\)
\(\frac{2}{5}x=\frac{4}{15}\)
\(x=\frac{2}{5}-\frac{4}{15}=>x=\frac{2}{15}\)
Vậy \(x=\frac{2}{15}\)
I) \(\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
\(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}\)
\(\frac{3}{5}x=\frac{5}{14}\)
\(x=\frac{5}{14}:\frac{3}{5}=\frac{5}{14}.\frac{5}{3}=\frac{25}{42}\)
Vậy \(x=\frac{25}{42}\)
21 tháng 12 2019
\(a)\frac{3}{4}+\frac{2}{5}x=\frac{29}{60}\)
\(\)TỰ LÀM NHA HIHI
22 tháng 12 2019
MI SUỐT NGÀY NGỒI MÁY TÍNH LƯỚT FACE, LÚC NÀO ĐI QUA CŨNG THẤY
CK
Tìm x biết :
\(\frac{1}{1×2×3×4}+\frac{1}{2×3×4×5}+\frac{1}{3×4×5×6}+...+\frac{1}{27×28×29×30}×x=-3\)
0
11 tháng 4 2019
\(1,-\frac{3}{29}+\frac{-7}{29}\le\frac{x}{29}\le-\frac{3}{29}-\frac{5}{29}\)
\(\Rightarrow-\frac{10}{29}\le\frac{x}{29}\le-\frac{8}{29}\Rightarrow-10\le x\le-8\)
\(\Rightarrow x=\left\{-8;-9;-10\right\}\)
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)
\(\Rightarrow2S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)
\(\Rightarrow2S-S=S=1-\frac{1}{2^{100}}\)
KA
3
26 tháng 5 2017
Đặt x + 29 = a (a \(\ne-29;-30\))
Đề trở thành: \(\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}=\frac{5}{4}\)
\(\Leftrightarrow\frac{\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}=\frac{5}{4}\)
\(\Leftrightarrow\frac{a^2+2a+1+a^2}{a^2\left(a^2+2a+1\right)}=\frac{5}{4}\)
\(\Leftrightarrow\frac{2a^2+2a+1}{a^4+2a^3+a^2}=\frac{5}{4}\)
\(\Leftrightarrow8a^2+8a+4=5a^4+10a^3+5a^2\)
\(\Leftrightarrow5a^4+10a^3-3a^2-8a-4=0\)
\(\Leftrightarrow5a^4+10a^3-3a^2-6a-2a-4=0\)
\(\Leftrightarrow5a^3\left(a+2\right)-3a\left(a+2\right)-2\left(a+2\right)=0\)
\(\Leftrightarrow\left(a+2\right)\left(5a^3-3a-2\right)=0\)
\(\Leftrightarrow\left(a+2\right)\left(5a^3-5a+2a-2\right)=0\)
\(\Leftrightarrow\left(a+2\right)\left(a-1\right)\left(5a^2+5a+2\right)=0\)
tới đây dễ r`
7 tháng 8 2017
\(\frac{3}{4}\)- \(\frac{2}{15}\). X =\(\frac{29}{60}\)
\(\frac{2}{15}\). X = \(\frac{29}{60}\)- \(\frac{45}{60}\)= \(\frac{-4}{15}\)
X = \(\frac{-4}{15}\). \(\frac{15}{2}\)
X = \(-2\)
Nhớ k cho mk nha
DG
7 tháng 8 2017
\(\frac{3}{4}\)- \(\frac{2}{15}.x=\frac{29}{60}\)
\(\frac{2}{15}.x=\frac{29}{60}+\frac{3}{4}\)
\(\frac{2}{15}.x=\frac{30}{37}\)
\(x=\frac{30}{37}:\frac{2}{15}\)
\(x=\frac{225}{37}\)
- đúng thì k giùm cái hen!!!!!!!!!