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bit lm bài này k giup tui
\(\frac{\left(\frac{518}{19}-\frac{342}{13}\right).\left(\frac{177}{236}+\frac{76}{236}-\frac{6}{236}\right)}{\left(\frac{3}{4}+x\right).\frac{27}{33}}=1\)
=>\(\frac{\left(\frac{6734}{247}-\frac{6498}{247}\right).\frac{247}{236}}{\left(\frac{3}{4}+x\right).\frac{27}{33}}=1\)
=>(3/4+x)*27/33=236/247*247/236=1
3/4+x=1:27/33=33/27
x=33/27-3/4=132/108-81/108
x=51/108
Vậy x=51/108
ta có : ( -5/28 +7/4 + 8/35 ) : (- 69/20)
= ( -25/140 + 245/140 + 32/140 ) x (-20/69)
= (252/140) x (-20/69)
= (9/5) x (-20/69)
= (- 12/23)
tính nhanh:
2 x 3/7 + (2/9 - 10/7) - 5/3 x 9
= 6/7 + 2/9 - 10/7 - 5/3 x 9 = 6/7 + 2/9 - 10/7 - 15
= (6/7 - 10/7 ) + (2/9 - 135/9) = ( - 4/7 ) + (-133/9 )
= (- 36/63) + (-931/63)
= (- 967/63)
Theo đề
=> \(\left|2x-1\right|-\frac{1}{2}=\frac{4}{5}\) hoặc \(\left|2x-1\right|-\frac{1}{2}=-\frac{4}{5}\)
=> |2x - 1| = 13/10 hoặc |2x - 1| = -3/10 (vô lí, loại)
=> 2x - 1 = 13/10 hoặc 2x - 1 = -13/10
=> 2x = 23/10 hoặc 2x = -3/10
=> x = 23/20 hoặc x = -3/20
Vậy...
\(\left|\left|2x-1\right|-\frac{1}{2}\right|=\frac{4}{5}\)
TH1 : \(\left|2x-1\right|-\frac{1}{2}=\frac{4}{5}\Rightarrow\left|2x-1\right|=\frac{13}{10}\)
TH2 : \(\left|2x-1\right|-\frac{1}{2}=-\frac{4}{5}\Rightarrow\left|2x-1\right|=\frac{-3}{10}\) (loại )
Ta có :
\(\left|2x-1\right|=\frac{13}{10}\)
=> TH1 : \(2x-1=\frac{13}{10}\Rightarrow2x=\frac{23}{10}\Rightarrow x=\frac{23}{20}\)
TH2 : \(2x-1=\frac{-13}{10}\Rightarrow2x=\frac{-3}{10}\Rightarrow x=\frac{-3}{20}\)
Vậy x = \(\frac{23}{20}\)
hoặc x = \(\frac{-3}{20}\)
Câu 1 :
Đk: \(x\ge1\)
\(\sqrt{x-1}+\sqrt{2x-1}=5\\ \Leftrightarrow x-1+2\sqrt{\left(x-1\right)\left(2x-1\right)}+2x-1=25\\ \Leftrightarrow2\sqrt{2x^2-3x+1}=27-3x\\ \)
\(\Leftrightarrow\begin{cases}27-3x\ge0\\4\left(2x^2-3x+1\right)=9x^2-162x+729\end{cases}\) \(\Leftrightarrow\begin{cases}x\le9\\x^2-150x+725=0\end{cases}\)
\(\Leftrightarrow\begin{cases}x\le9\\x=145hoặcx=5\end{cases}\)
với x= 5 thoản mãn điều kiện, x=145 loại
Vậy \(S=\left\{5\right\}\)
đk: \(\begin{cases}x+2\ne0\\4-x>0\\6+x>0\end{cases}\)
ta có \(3\log_{\frac{1}{4}}\left(x+2\right)-3=3\log_{\frac{1}{4}}\left(4-x\right)+3\log_{\frac{1}{4}}\left(6+x\right)\) suy ra \(\log_{\frac{1}{4}}\left(x+2\right)-\log_{\frac{1}{4}}\frac{1}{4}=\log_{\frac{1}{4}}\left(4-x\right)\left(6+x\right)\) suy ra \(\log_{\frac{1}{4}}\left(x+2\right).\frac{1}{4}=\log_{\frac{1}{4}}\left(4-x\right)\left(6+x\right)\) suy ra \(\frac{x+2}{4}=\left(4-x\right)\left(6+x\right)\)
giải pt tìm ra x
đối chiếu với đk của bài ta suy ra đc nghiệm của pt
\(\frac{1}{3}+\frac{1}{2.3}\left(1+2\right)+\frac{1}{3.3}\left(1+2+3\right)+...+\frac{1}{3.2015}\left(1+2+3+...+2015\right)=\frac{1}{3}\left[\frac{2}{2}+\frac{1}{2}\left(\frac{2.3}{2}\right)+\frac{1}{3}\left(\frac{3.4}{2}\right)+...+\frac{1}{2015}\left(\frac{2016.2015}{2}\right)\right]=\frac{1}{3}.\frac{1}{2}\left(2+3+4+....+2016\right)=\frac{1}{6}\left(\frac{2016.2017}{2}-1\right)\)
2x-\(\frac{1}{3}\)=1-\(\frac{5}{6}\)
2x-\(\frac{1}{3}\)=\(\frac{1}{6}\)
2x=\(\frac{1}{6}\)+\(\frac{1}{3}\)
2x=1/6 +2/6
2x=\(\frac{1}{2}\)
x=1/2 : 2
x/\(\frac{1}{4}\)
\(\frac{7}{9}\):(2+\(\frac{3}{4}\)x)+\(\frac{5}{9}\)=\(\frac{23}{27}\)
7/9 :(2+3/4x)=\(\frac{23}{27}\)-\(\frac{5}{9}\)
7/9 :(2+3/4x)=\(\frac{23}{27}\)-\(\frac{15}{27}\)
7/9 :(2+3/4x)=\(\frac{8}{27}\)
(2+3/4x) =\(\frac{7}{9}\) . \(\frac{27}{8}\)
(2+3/4x) =\(\frac{21}{8}\)
\(\frac{3}{4}\)x =\(\frac{21}{8}\)-2
3/4x =21/8 -16/8
3/4x = 5/8
x =\(\frac{5}{8}\) : \(\frac{3}{4}\)
x =5/8 . 4/3
x =\(\frac{20}{24}\)