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\(-\frac{5}{9}\left(\frac{3}{10}-\frac{2}{5}\right)=-\frac{5}{9}\left(\frac{3}{10}-\frac{4}{10}\right)=-\frac{5}{9}.\frac{-1}{10}=\frac{1}{18}\)
\(\frac{1}{2}\sqrt{64}-\sqrt{\frac{9}{25}}+1^{2016}=\frac{1}{2}.8-\frac{3}{5}+1=4+\frac{2}{5}=\frac{22}{5}\)
\(2^8:2^5+3^2.2-12=2^3+9.2-12=8+18-12=8+6=14\)
\(3^x+\sqrt{\frac{16}{81}}-\sqrt{9}+\frac{\sqrt{81}}{3}=9\frac{4}{9}\)
\(3^x+\frac{4}{9}-3+\frac{9}{3}=9\frac{4}{9}\)
\(3^x+\frac{4}{9}-3+3=9\frac{4}{9}\)
\(3^x+\frac{4}{9}=9+\frac{4}{9}\)
\(\Rightarrow3^x=9+\frac{4}{9}-\frac{4}{9}\)
\(3^x=9\)
\(3^x=3^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
a, \(\frac{-5}{9}.\left(\frac{3}{10}-\frac{2}{5}\right)\)
\(=\frac{-5}{9}.\left(\frac{3}{10}-\frac{4}{10}\right)\)
\(=\frac{-5}{9}.\frac{-1}{10}\)
\(=\frac{5}{90}\)
\(=\frac{1}{18}\)
b,\(\frac{2}{3}+\frac{-1}{3}+\frac{7}{15}\)
\(=\frac{10}{15}-\frac{5}{15}+\frac{7}{15}\)
\(=\frac{12}{15}\)
\(=\frac{4}{5}\)
c, \(\frac{3}{8}.3\frac{1}{3}\)
\(=\frac{3}{8}.\frac{10}{3}\)
\(=\frac{10}{8}\)
\(=\frac{5}{4}\)
d, \(\frac{-3}{5}+0,8.\left(-7\frac{1}{2}\right)\)
\(=\frac{-3}{5}+\frac{4}{5}.\frac{-15}{2}\)
\(=\frac{-3}{5}+\frac{-60}{10}\)
\(=\frac{-3}{5}+\frac{-30}{5}\)
\(=\frac{-33}{5}\)
e, \(\frac{2}{5}.8\frac{1}{3}+1\frac{2}{3}.\frac{2}{5}\)
\(=\frac{2}{5}.\left(8\frac{1}{3}+1\frac{2}{3}\right)\)
\(=\frac{2}{5}.10\)
\(=4\)
f, \(\frac{3}{7}.19\frac{1}{3}-\frac{3}{7}.33\frac{1}{3}\)
\(=\frac{3}{7}.\left(19\frac{1}{3}-33\frac{1}{3}\right)\)
\(=\frac{3}{7}.-14\)
\(=-6\)
~Study well~
#KSJ
\(\left|2x-27\right|^{2007}+\left(3y+10\right)^{2018}=0\)
Ta có \(\left|2x-27\right|^{2017}\ge0\forall x;\left(3y+10\right)^{2018}\ge0\forall y\)
\(\Rightarrow\left|2x-27\right|^{2017}+\left(3.y+10\right)^{2018}\ge0\forall x;y\)
\(\Rightarrow\left|2x-17\right|^{2017}+\left(3y+10\right)^{2018}=0\)
\(\Leftrightarrow\hept{\begin{cases}2x-17=0\\3.y+10=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{17}{2}\\y=-\frac{10}{3}\end{cases}}\)
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) \(\left(\frac{2}{3}x-\frac{1}{2}\right)\cdot\frac{-2}{3}+\frac{1}{5}=\frac{-3}{4}\)
=> \(\left(\frac{2}{3}x-\frac{1}{2}\right)\cdot\frac{-2}{3}=-\frac{19}{20}\)
=> \(\frac{2}{3}x-\frac{1}{2}=\left(-\frac{19}{20}\right):\left(-\frac{2}{3}\right)=\left(-\frac{19}{20}\right)\cdot\left(-\frac{3}{2}\right)=\frac{57}{40}\)
=> \(\frac{2}{3}x=\frac{57}{40}+\frac{1}{2}=\frac{77}{40}\)
=> \(x=\frac{77}{40}:\frac{2}{3}=\frac{77}{40}\cdot\frac{3}{2}=\frac{231}{80}\)
1.
a)\(\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
b)\(\left(x-2\right)^2=1\Leftrightarrow\orbr{\begin{cases}x-2=\sqrt{1}\\x-2=-\sqrt{1}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{1}+2\\x=-\sqrt{1}+2\end{cases}}\)
Mấy câu kia tương tự,bạn tự làm nha :))
a/ Ta có \(\left|\frac{5}{6}-2x\right|=\frac{7}{8}\)
=> \(\orbr{\begin{cases}\frac{5}{6}-2x=\frac{7}{8}\\\frac{5}{6}-2x=\frac{-7}{8}\end{cases}}\)=> \(\orbr{\begin{cases}-2x=\frac{1}{24}\\-2x=\frac{-41}{24}\end{cases}}\)=> \(\orbr{\begin{cases}x=-\frac{1}{48}\\x=\frac{41}{48}\end{cases}}\)
Vậy \(x=-\frac{1}{48}\)hoặc \(x=\frac{41}{48}\)thì \(\left|\frac{5}{6}-2x\right|=\frac{7}{8}\)
b/ Ta có \(B=5x^2-7y+6\)
Thay \(x=\frac{-1}{5}\)và \(y=\frac{-3}{7}\)vào biểu thức B, ta có:
\(5\left(-\frac{1}{5}\right)^2-7\left(-\frac{3}{7}\right)+6\)= \(\frac{1}{5}-\left(-3\right)+6=\frac{1}{5}+3+6=\frac{1}{5}+9=\frac{46}{5}\)
Vậy giá trị của biểu thức B bằng \(\frac{46}{5}\)khi \(x=\frac{-1}{5}\)và \(y=\frac{-3}{7}\).
a/ Ta có 6 5 − 2x = 8 7 => 6 5 − 2x = 8 7 6 5 − 2x = 8 −7 => −2x = 24 1 −2x = 24 −41
=> x = − 48 1 x = 48 41 Vậy x = − 48 1 hoặc x = 48 41 thì 6 5 − 2x = 8 7
b/ Ta có B = 5x 2 − 7y + 6 Thay x = 5 −1 và y = 7 −3 vào biểu thức B, ta có: 5 − 5 1 2 − 7 − 7 3 + 6= 5 1 − −3 + 6 = 5 1 + 3 + 6 = 5 1 + 9 = 5 46
Vậy giá trị của biểu thức B bằng 5 46 khi x = 5 −1 và y = 7 −3 .
a,.\(\frac{1}{2}+\frac{2}{3}.\sqrt{9}\)
=\(\frac{1}{2}+\frac{2}{3}.3\)
=\(\frac{1}{2}+2\)
=\(\frac{5}{2}\)
b,\(x+\frac{2}{5}=1\)
\(x=1-\frac{2}{5}\)
X =\(\frac{3}{5}\)
a, \(\frac{1}{2}+\frac{2}{3}\cdot\sqrt{9}\)
\(\Leftrightarrow\frac{1}{2}+\frac{2}{3}\cdot3\)
\(=\frac{1}{2}+2\)
\(=\frac{5}{2}\)
b, \(x+\frac{2}{5}=1\)
\(x=1+\left(\frac{-2}{5}\right)\)
\(x=\frac{3}{5}\)
c, Ta có : \(A=\left|x+\frac{1}{2}\right|\ge0\)
\(\Rightarrow A_{min}=0\)
Dấu "=" xảy ra khi : \(\left|x+\frac{1}{2}\right|=0\Leftrightarrow x=-\frac{1}{2}\)
Vậy \(A_{min}=0\Leftrightarrow x=-\frac{1}{2}\)
d, \(2x-5=0\Leftrightarrow2x=0+5\Leftrightarrow2x=5\Leftrightarrow x=\frac{5}{2}\)
Vậy tập hợp các nghiệm của phương trình \(2x-5=0\) là \(\left\{\frac{5}{2}\right\}\).