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Bài 1:
a) \(\frac{2}{5}+\frac{1}{5}.\left(\frac{3}{4}\right)\)
= \(\frac{2}{5}+\frac{3}{20}\)
= \(\frac{11}{20}\)
b) \(\frac{5}{12}.\left(-\frac{3}{4}\right)\) + \(\frac{7}{12}.\left(-\frac{3}{4}\right)\)
= \(\left(\frac{5}{12}+\frac{7}{12}\right).\left(-\frac{3}{4}\right)\)
= 1.\(\left(-\frac{3}{4}\right)\)
= \(-\frac{3}{4}\)
Còn câu c) đang nghĩ.
Bài 2:
a) \(\frac{5}{7}+\frac{2}{7}\)x = 1
1.x = 1
x = 1 : 1
x = 1
Vậy x = 1.
b) 0,2 + | x - 1, 3 = 1, 5|
0,2 + x = 1, 5 + 1, 3
0,2 + x = 2, 8
x = 2, 8 - 0, 2
x = 2, 6
Vậy x = 2, 6.
c) 2x + 5 = 37
2x = 37 - 5
2x = 32
2x = 25
=> x = 5
Vậy x = 5.
d) 2x + 2x + 1 = 48
2x . 1 + 2x . 21 = 48
2x . ( 1 + 2) = 48
2x . 3 = 48
2x = 48 : 3
2x = 16
2x = 24
=> x = 4
Vậy x = 4.
Chúc bạn học tốt!
làm bước trung gian giùm mình luôn nhé
thanks trước những bạn làm giùm nhé
mình đang cần gấp lắm sáng mai là mình cần ai đang on làm giùm mình nhé
thanks
Bạn ghi ra nhiều vậy người khác nhìn rối mắt không trả lời được đâu ghi từng bài ra thôi
Mình chỉ làm được vài bài thôi, kiến thức có hạn :>
Bài 1:
Câu a và c đúng
Bài 2:
a) |x| = 2,5
=>x = 2,5 hoặc
x = -2,5
b) |x| = 0,56
=>x = 0,56
x = - 0,56
c) |x| = 0
=. x = 0
d)t/tự
e) |x - 1| = 5
=>x - 1 = 5
x - 1 = -5
f) |x - 1,5| = 2
=>x - 1,5 = 2
x - 1,5 = -2
=>x = 2 + 1,5
x = -2 + 1,5
=>x = 3,5
x = - 0,5
các câu sau cx t/tự thôi
Bài 3: Ko hỉu :)
Bài 4: Kiến thức có hạn :)
a.3 - | x + 7 | - 1/2 = 1/3
3 - | x + 7 | = 1/3 +1/2
3 - | x +7 | = 5/6
| x+ 7 | = 3 - 5/6
| x + 7| = 13/ 6
roi chia thanh 2 truong hop la xong ok
a) \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)
=> \(\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}\)
=> \(\frac{2}{3}x=-\frac{29}{70}\)
=> \(x=-\frac{29}{70}:\frac{2}{3}\)
=> \(x=-\frac{29}{70}.\frac{3}{2}\)
=> \(x=-\frac{87}{140}\)
b) \(-\frac{21}{13}x+\frac{1}{3}=-\frac{2}{3}\)
=> \(-\frac{21}{13}x=-\frac{2}{3}-\frac{1}{3}\)
=> \(-\frac{21}{13}x=-\frac{3}{3}\)
=> \(-\frac{21}{13}x=1\)
=> \(x=1:\left(-\frac{21}{13}\right)\)
=> \(x=-\frac{13}{21}\)
c) \(\left|x-1,5\right|=2\)
=> \(\left[{}\begin{matrix}x-1,5=2\\x-1,5=-2\end{matrix}\right.=>\left[{}\begin{matrix}x=2+1,5\\x=-2+1,5\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.=>\left[{}\begin{matrix}x=\frac{7}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)(T/M)
d) \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
=> \(\left|x+\frac{3}{4}\right|=\frac{1}{2}\)
=> \(=>\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{matrix}\right.=>\left[{}\begin{matrix}x=\frac{1}{2}-\frac{3}{4}\\x=-\frac{1}{2}-\frac{3}{4}\end{matrix}\right.=>\left[{}\begin{matrix}x=-\frac{1}{4}\\x=-\frac{5}{4}\end{matrix}\right.\)(T/M)
HỌC TỐT 
a) \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)
\(\Leftrightarrow\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}\)
\(\Leftrightarrow\frac{2}{3}x=-\frac{29}{70}\)
\(\Leftrightarrow x=-\frac{29}{70}:\frac{2}{3}\)
\(\Leftrightarrow x=-\frac{87}{140}\)
b) \(-\frac{21}{13}x+\frac{1}{3}=-\frac{2}{3}\)
\(\Leftrightarrow-\frac{21}{13}x=-\frac{2}{3}-\frac{1}{3}\)
\(\Leftrightarrow-\frac{21}{13}x=-1\)
\(\Leftrightarrow x=-1:\left(-\frac{21}{13}\right)\)
\(\Leftrightarrow x=\frac{13}{21}\)
c) \(\left|x-1,5\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1,5=2\\x-1,5=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.\)
d) \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
\(\Leftrightarrow\left|x+\frac{3}{4}\right|=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{4}\\x=\frac{5}{4}\end{matrix}\right.\)
\(\frac{x}{11,9}=\frac{1,5}{3,5}\Rightarrow x=\frac{1,5.11,9}{3,5}\Rightarrow x=5,1\)
\(\frac{x}{0,2}=\frac{80}{x}\Rightarrow x^2=80.0,2\Rightarrow x^2=16\Rightarrow x=4;-4\)
\(\frac{2}{3}x:\frac{1}{5}=\frac{4}{3}:\frac{1}{4}\Rightarrow\frac{2}{3}x:\frac{1}{5}=\frac{16}{3}\Rightarrow\frac{2}{3}x=\frac{16}{15}\Rightarrow x=\frac{8}{5}\)
\(3:\frac{2}{5}x=\frac{1}{0,001}\Rightarrow\frac{2}{5}x=\frac{3}{1000}\Rightarrow x=\frac{3}{400}\)
\(\frac{x}{11,9}=\frac{1,5}{3,5}\Leftrightarrow x=\frac{11,9.1,5}{3,5}=5,1\)
Tương tự
a) Ta có: \(\frac{1}{2}+\frac{2}{3}:\left(x-1\right)=\frac{2}{3}\)
⇒\(\frac{2}{3}:\left(x-1\right)=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}\)
⇒\(x-1=\frac{2}{3}:\frac{1}{6}=\frac{2}{3}\cdot6=4\)
hay x=5
Vậy: x=5
b) \(5,4-3\left[x-120\%\right]=\frac{3}{10}\)
⇔\(\frac{27}{5}-3\cdot\left(x-\frac{6}{5}\right)=\frac{3}{10}\)
⇔\(3\left(x-\frac{6}{5}\right)=\frac{27}{5}-\frac{3}{10}=\frac{51}{10}\)
hay \(x-\frac{6}{5}=\frac{51}{10}\cdot\frac{1}{3}=\frac{17}{10}\)
⇔\(x=\frac{17}{10}+\frac{6}{5}=\frac{29}{10}\)
Vậy: \(x=\frac{29}{10}\)
c) \(10\cdot3^{x+2}-3^x=89\)
\(\Leftrightarrow10\cdot3^2\cdot3^x-3^x=89\)
\(\Leftrightarrow3^x\left(90-1\right)=89\)
\(\Leftrightarrow3^x=1\)
hay x=0
Vậy: x=0
d) \(5\cdot\left(x-0,2\right)=3x+\left(\frac{-2}{3}\right)^3\)
⇒\(5\cdot\left(x-\frac{1}{5}\right)=3x+\frac{-8}{27}\)
\(\Leftrightarrow5x-1-3x-\frac{-8}{27}=0\)
\(\Leftrightarrow2x-\frac{19}{27}=0\)
\(\Leftrightarrow2x=\frac{19}{27}\)
hay \(x=\frac{\frac{19}{27}}{2}=\frac{19}{27}\cdot\frac{1}{2}=\frac{19}{54}\)
Vậy: \(x=\frac{19}{54}\)
e) \(\left(2x+\frac{3}{4}\right)^2-1,5=2\frac{1}{2}\)
\(\Leftrightarrow\left(2x+\frac{3}{4}\right)^2=\frac{5}{2}+\frac{3}{2}=\frac{8}{2}=4\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{3}{2}=-2\\2x+\frac{3}{2}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-2-\frac{3}{2}\\2x=2-\frac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\frac{7}{2}\\2x=\frac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-7}{2}\cdot\frac{1}{2}\\x=\frac{1}{2}\cdot\frac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-7}{4}\\x=\frac{1}{4}\end{matrix}\right.\)
Vậy: \(x\in\left\{-\frac{7}{4};\frac{1}{4}\right\}\)
a) \(\frac{3}{4}+\frac{1}{4}.x=\frac{1}{2}+\frac{1}{2}x\)
\(\Rightarrow3.\frac{1}{4}+\frac{1}{4}.x=\frac{1}{2}.\left(x+1\right)\)
\(\Rightarrow\frac{1}{4}.\left(x+3\right)=\frac{1}{2}.\left(x+1\right)\)
\(\Rightarrow\frac{x+1}{x+3}=\frac{1}{4}:\frac{1}{2}=\frac{1}{2}\)\(\Rightarrow\left(x+1\right).2=x+3\Rightarrow2x+2=x+3\)
\(\Rightarrow2x-x=3-2\Rightarrow x=1\)
vay x=1
a) \(\frac{3}{5}.x-\frac{1}{5}=\frac{4}{5}\)
\(\Leftrightarrow\frac{3}{5}.x=\frac{4}{5}+\frac{1}{5}\)
\(\Leftrightarrow\frac{3}{5}.x=1\)
\(\Leftrightarrow x=1:\frac{3}{5}\)
\(\Leftrightarrow x=\frac{5}{3}\)
Vậy : \(x=\frac{5}{3}\)
b) \(\frac{4}{7}+\frac{5}{7}:x=1\)
\(\Leftrightarrow\frac{5}{7}:x=1-\frac{4}{7}\)
\(\Leftrightarrow\frac{5}{7}:x=\frac{3}{7}\)
\(\Leftrightarrow x=\frac{5}{7}:\frac{3}{7}\)
\(\Leftrightarrow x=\frac{5}{3}\)
Vậy : \(x=\frac{5}{3}\)
c) \(-\frac{12}{7}.\left(\frac{3}{4}-x\right).\frac{1}{4}=-1\)
\(\Leftrightarrow\frac{-12.1}{7.4}.\left(\frac{3}{4}-x\right)=-1\)
\(\Leftrightarrow-\frac{3}{7}.\left(\frac{3}{4}-x\right)=-1\)
\(\Leftrightarrow\frac{3}{4}-x=-1:\left(-\frac{3}{7}\right)\)
\(\Leftrightarrow\frac{3}{4}-x=\frac{7}{3}\)
\(\Leftrightarrow x=\frac{3}{4}-\frac{7}{3}=-\frac{19}{12}\)
Vậy : \(x=-\frac{19}{12}\)
d) \(x:\frac{17}{8}=-\frac{2}{5}.-\frac{9}{17}+3\)
\(\Leftrightarrow x:\frac{17}{8}=\frac{273}{85}\)
\(\Leftrightarrow x=\frac{273}{85}.\frac{17}{8}\)
\(\Leftrightarrow x=\frac{273}{40}\)
Vậy : \(x=\frac{273}{40}\)
\(\)
a, \(x.\left(\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^3\)
\(\Rightarrow x=\left(\frac{3}{5}\right)^3:\left(\frac{3}{5}\right)^2\)
\(\Rightarrow x=\frac{3}{5}\)
b, \(3,5+\left|x+\frac{3}{2}\right|=-1,5\)
Ta có: \(3,5+\left|x+\frac{3}{2}\right|>0,-1,5< 0\)
Nên không có x thỏa mãn
c, \(3^{x+2}+3^x=270\)
\(\Rightarrow3^x\left(3^2+1\right)=270\)
\(\Rightarrow3^x.10=270\)
\(\Rightarrow3^x=27\Rightarrow3^x=3^3\Rightarrow x=3\)
Chúc bạn học tốt.