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ừ Vy Nguyễn, mik làm nè:
e, \(\dfrac{-2}{3}-\dfrac{1}{3}\left(2x-5\right)=\dfrac{3}{2}.\)
\(\dfrac{1}{3}\left(2x-5\right)=\dfrac{-2}{3}-\dfrac{3}{2}.\)
\(\dfrac{1}{3}\left(2x-5\right)=\dfrac{-4}{6}+\dfrac{-9}{6}.\)
\(\dfrac{1}{3}\left(2x-5\right)=\dfrac{-13}{6}.\)
\(2x-5=\dfrac{-13}{6}:\dfrac{1}{3}.\)
\(2x-5=\dfrac{-13}{6}.3.\)
\(2x-5=\dfrac{-13}{2}.\)
\(2x=\dfrac{-13}{2}+5.\)
\(2x=\dfrac{-13}{2}+\dfrac{10}{2}.\)
\(2x=\dfrac{-3}{2}.\)
\(x=\dfrac{-3}{2}:2.\)
\(x=\dfrac{-3}{2.2}=\dfrac{-3}{4}.\)
g, \(\dfrac{2}{5}x+\dfrac{1}{2}=\dfrac{-3}{4}.\)
\(\dfrac{2}{5}x=\dfrac{-3}{4}-\dfrac{1}{2}.\)
\(\dfrac{2}{5}x=\dfrac{-3}{4}+\dfrac{-2}{4}.\)
\(\dfrac{2}{5}x=\dfrac{-5}{4}.\)
\(x=\dfrac{-5}{4}:\dfrac{2}{5}.\)
\(x=\dfrac{-5}{4}.\dfrac{5}{2}.\)
\(x=\dfrac{-25}{8}.\)
h, \(\left(2x-2\dfrac{4}{5}\right):3\dfrac{1}{8}=1\dfrac{3}{5}.\)
\(\left(2x-2\dfrac{4}{5}\right)=\dfrac{8}{5}.\dfrac{25}{8}.\)
\(\left(2x-2\dfrac{4}{5}\right)=5.\)
\(2x=5+2\dfrac{4}{5}.\)
\(2x=7\dfrac{4}{5}.\)
\(x=7\dfrac{4}{5}:2.\)
\(x=\dfrac{39}{10}.\)
(còn tiếp ở phần sau!!!)
Tiếp:
i, \(3,2x-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):3\dfrac{2}{3}=\dfrac{7}{20}.\)
\(\dfrac{16}{5}x-\left(\dfrac{4}{5}+\dfrac{2}{3}\right)=\dfrac{7}{20}.\dfrac{11}{3}.\)
\(\dfrac{16}{5}x-\left(\dfrac{4}{5}+\dfrac{2}{3}\right)=\dfrac{77}{60}.\)
\(\dfrac{16}{5}x-\left(\dfrac{12}{15}+\dfrac{10}{15}\right)=\dfrac{77}{60}.\)
\(\dfrac{16}{5}x-\dfrac{22}{15}=\dfrac{77}{60}.\)
\(\dfrac{16}{5}x=\dfrac{77}{60}+\dfrac{22}{15}.\)
\(\dfrac{16}{5}x=\dfrac{77}{60}+\dfrac{88}{60}.\)
\(\dfrac{16}{5}x=\dfrac{165}{60}=\dfrac{11}{4}.\)
\(x=\dfrac{11}{4}:\dfrac{16}{5}.\)
\(x=\dfrac{11}{4}.\dfrac{5}{16}=\dfrac{55}{64}.\)
k, \(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}.\)
\(\left(\dfrac{3x}{7}+1\right)=\dfrac{-1}{28}.\left(-4\right).\)
\(\left(\dfrac{3x}{7}+1\right)=\dfrac{1}{7}.\)
\(\dfrac{3x}{7}=\dfrac{1}{7}-1.\)
\(\dfrac{3x}{7}=\dfrac{1}{7}-\dfrac{7}{7}.\)
\(\dfrac{3x}{7}=\dfrac{-6}{7}.\)
\(\Rightarrow3x=-6.\)
\(\Rightarrow x=-6:3=-2.\)
~ Chúc bn học tốt!!! ~
Bài mik đúng thì nhớ tik mik nha!!!
a)\(\dfrac{-1}{3}+\dfrac{2}{1}-\dfrac{6}{5}=\dfrac{-5}{15}+\dfrac{30}{15}-\dfrac{18}{15}=\dfrac{7}{15}\)
dai dong qua(de)
3/ Chu vi hình chữ nhật:
\(\left(\dfrac{1}{4}+\dfrac{3}{10}\right)\cdot2=\dfrac{11}{10}\) (chưa biết đơn vị)
Diện tích hình chữ nhật:
\(\dfrac{1}{4}\cdot\dfrac{3}{10}=\dfrac{11}{20}\) (chưa biết đơn vị)
a) \(\left|x-1\right|-1=2\)
\(\Rightarrow\left|x-1\right|=3\Rightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
Vậy......
b) \(\left|5x+1\right|+\left|6y-3\right|\le0\)
Vì \(\left\{{}\begin{matrix}\left|5x+1\right|\ge0\forall x\\\left|6y-3\right|\ge0\forall y\end{matrix}\right.\) Để biểu thức <= 0
\(\Rightarrow\left\{{}\begin{matrix}\left|5x+1\right|=0\\\left|6y-3\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{5}\\y=\dfrac{1}{2}\end{matrix}\right.\)
Vậy........
c) \(\left|3x-1\right|+\left(2y-1\right)^{20}=0\)
Vì \(\left\{{}\begin{matrix}\left|3x-1\right|\ge0\forall x\\\left(2y-1\right)^{20}\ge0\forall y\end{matrix}\right.\)
Để biểu thức = 0
\(\Rightarrow\left\{{}\begin{matrix}\left|3x-1\right|=0\\\left(2y-1\right)^{20}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=\dfrac{1}{2}\end{matrix}\right.\)
Vậy........
d/ \(\left|x-3\right|+\left|x+10\right|=13\)
a. \(\left|x-1\right|=3\)
=> x-1 = 3 hoặc x-1 = -3
=> x = 4 hoặc x = -2
a) \(\left(x-3\right)^3=27\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^3=3^3\\\left(x-3\right)^3=\left(-3\right)^3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-3=3\\x-3=-3\end{matrix}\right.\)
\(\)\(\Rightarrow\left[{}\begin{matrix}x=6\\x=0\end{matrix}\right.\) \(\left(TM\right)\)
Vậy \(x\in\left\{0;6\right\}\) là giá trị cần tìm
b) \(\left(2x+1\right)^3=125\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x+1\right)^3=5^3\\\left(2x+1\right)^3=\left(-5\right)^3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\) \(\left(TM\right)\)
Vậy \(x\in\left\{3;-2\right\}\) là giá trị cần tìm
c) \(2^x-15=17\)
\(2^x=17+15\)
\(2^x=32\)
\(2^x=2^5\)
\(\Rightarrow x=5\left(TM\right)\)
Vậy \(x=5\) là giá trị cần tìm
\(\left(x-3\right)^3=27\)
\(\Leftrightarrow\left(x-3\right)^3=3^3\)
\(\Leftrightarrow x-3=3\Rightarrow x=6\)
\(\left(2x+1\right)^3=125\)
\(\Leftrightarrow\left(2x+1\right)^3=5^3\)
\(\Leftrightarrow2x+1=5\Rightarrow2x=4\Rightarrow x=2\)
\(2^x-15=17\)
\(\Leftrightarrow2^x=15+17=32\)
\(\Leftrightarrow x=5\)
\(\left(7x-11\right)^3=25.3^2+20\)
\(\left(7x-11\right)^3=245\)
\(\left(7x-11\right)^3=\sqrt[3]{245}\)
(lớp 6 chưa học cái này đề sai nhé)
\(2^{x+3}+2^x=36\)
\(2^x.2^3+2^x=36\)
\(2^x\left(2^3+1\right)=36\)
\(2^x.9=36\Rightarrow2^x=4\Rightarrow x=2\)
\(3^{x+4}+4+3^{x+2}=270\)
\(3^{x+4}+3^{x+2}=270-4=266\)
\(3^x.3^4+3^x.3^2=266\)
\(3^x\left(3^4+3^2\right)=266\)(đề cũng sai)
\(2.3^x=10.3^{12}+8.27^4\)
\(2.3^x=5314410+4251528\)
\(2.3^x=9565938\)
\(3^x=9565938:2=4782969\)
\(\Leftrightarrow x=14\)
\(\left(2x-5\right)^5=3^{10}\)
\(\Leftrightarrow\left(2x-5\right)^5=\left(3^2\right)^5\)
\(\left(2x-5\right)^5=9^5\)
\(\Leftrightarrow2x-5=9\Rightarrow2x=14\Rightarrow x=7\)