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Bài 2.
A = -3/5 + ( -2/5 + 2 )
A = -3/5 + ( -2/5 + 10/5 )
A = -3/5 + 8/5
A = 5/5
A = 1
--------------------------------------------------------
B = 3/7 + ( -1/5 + -3/7 )
B = 3/7 + ( -7/35 + -15/35 )
B = 3/7 + ( -22/35 )
B = 15/35 + ( -22/35 )
B = -1/5
-----------------------------------------------------
C = ( -5/24 + 0,75 + 7/12 ) : ( -2 . 1/8 )
C = ( -5/24 + 3/4 + 7/12 ) : ( -1/4 )
C = 9/8 : ( -1/4 )
C = 9/8 . ( -4 )
C = -9/2
Bài 3 .
a) 4/7 - x = 1/2 . x + 2/7
<=> -x - x = 1/2 - 4/7 + 2/7
<=> -2x = 3/14
<=> x = 3/14 . ( -1/2 )
<=> x = -3/28
Vậy x = -3/28
b) x : 3 1/5 = 1 1/2
<=> x : 16/5 = 3/2
<=> x = 3/2 . 16/5
<=> x = 24/5
Vậy x = 24/5
c) x . 3/4 = -1 5/8
<=> x . 3/4 = -13/8
<=> x = -13/8 . 4/3
<=> x = -13/6
Vậy x = -13/6
Bài 1 : Rút gọn các phân số sau đến tối giản :
a) \(\dfrac{3.21}{14.15}=\dfrac{3.3.7}{2.7.3.5}=\dfrac{1.3.1}{2.1.1.5}=\dfrac{3}{10}\)
b) \(\dfrac{49+49.7}{49}=\dfrac{49\left(1+7\right)}{49}=\dfrac{49.8}{49}=\dfrac{1.8}{1}=\dfrac{8}{1}=8\)
A= \(\dfrac{-3}{5}-\dfrac{-4}{5}+\dfrac{-9}{10}\)
A = \(\dfrac{-7}{10}\)
\(\left(x-2\right)\left(y+3\right)=5\)
\(\Rightarrow x-2;y+3\inƯ\left(5\right)\)
\(Ư\left(5\right)=\left\{\pm1;\pm5\right\}\)
Xét ước
\(xy-6x-3y=7\)
\(\Rightarrow xy-6x-3y+18=25\)
\(\Rightarrow x\left(y-6\right)-3\left(y-6\right)=25\)
\(\Rightarrow\left(x-3\right)\left(y-6\right)=25\)
Xét ước
\(\dfrac{a}{2}-\dfrac{1}{b}=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{b}=\dfrac{3}{4}+\dfrac{a}{2}\)
\(\Rightarrow\dfrac{1}{b}=\dfrac{3}{4}+\dfrac{2a}{4}\)
\(\Rightarrow\dfrac{1}{b}=\dfrac{3+2a}{4}\)
\(\Rightarrow b\left(3+2a\right)=4\)
Xét ước
3. Gọi d là ƯCLN(2n + 3, 4n + 8), d ∈ N*
\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\4n+8⋮d\end{cases}\Rightarrow\hept{\begin{cases}2\left(2n+3\right)⋮d\\4n+8⋮d\end{cases}\Rightarrow}\hept{\begin{cases}4n+6⋮d\\4n+8⋮d\end{cases}}}\)
\(\Rightarrow\left(4n+8\right)-\left(4n+6\right)⋮d\)
\(\Rightarrow2⋮d\)
\(\Rightarrow d\in\left\{1;2\right\}\)
Mà 2n + 3 không chia hết cho 2
\(\Rightarrow d=1\)
\(\RightarrowƯCLN\left(2n+3,4n+8\right)=1\)
\(\Rightarrow\frac{2n+3}{4n+8}\) là phân số tối giản.
Theo đề bài ta có \(\dfrac{a}{b}=\dfrac{c}{d}\)
=> \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\) ( tính chất dãy tỉ số = nhau )
=> \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a-c}{b-d}\) ( tính chất dãy tỉ số = nhau )
Câu 2:
Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a+b}{b}=\dfrac{bk+b}{b}=k+1\)
\(\dfrac{c+d}{d}=\dfrac{dk+d}{d}=k+1\)
Do đó: \(\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
Bài 1:
a)
\(\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{x-1}{9}=\dfrac{24}{9}\\ \Leftrightarrow x-1=24\\ x=24+1\\ x=25\)
b)
\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{8}\\ \dfrac{3x}{7}+1=\dfrac{-1}{8}\cdot\left(-4\right)\\ \dfrac{3x}{7}+1=\dfrac{1}{2}\\ \dfrac{3x}{7}=\dfrac{1}{2}-1\\ \dfrac{3x}{7}=\dfrac{-1}{2}\\ 3x=\dfrac{-1}{2}\cdot7\\ 3x=\dfrac{-7}{2}\\ x=\dfrac{-7}{2}:3\\ x=\dfrac{-7}{6}\)
c)
\(x+\dfrac{7}{12}=\dfrac{17}{18}-\dfrac{1}{9}\\ x+\dfrac{7}{12}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{12}\\ x=\dfrac{1}{4}\)
d)
\(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\\ \dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=\dfrac{7}{12}\\ \dfrac{-1}{6}x=\dfrac{7}{12}\\ x=\dfrac{7}{12}:\dfrac{-1}{6}\\ x=\dfrac{-7}{2}\)
e)
\(\dfrac{29}{30}-\left(\dfrac{13}{23}+x\right)=\dfrac{7}{46}\\ \dfrac{29}{30}-\dfrac{13}{23}-x=\dfrac{7}{46}\\ \dfrac{277}{690}-x=\dfrac{7}{46}\\ x=\dfrac{277}{690}-\dfrac{7}{46}\\ x=\dfrac{86}{345}\)
f)
\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\\ \left(x-\dfrac{1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\\ x-\dfrac{1}{12}=\dfrac{7}{46}\cdot\dfrac{23}{12}\\ x-\dfrac{1}{12}=\dfrac{7}{24}\\ x=\dfrac{7}{24}+\dfrac{1}{12}\\ x=\dfrac{3}{8}\)
g)
\(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{1}{6}\\ \dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\\ \dfrac{13}{21}+x=\dfrac{2}{7}\\ x=\dfrac{2}{7}-\dfrac{13}{21}\\ x=\dfrac{-1}{3}\)
h)
\(2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\end{matrix}\right.\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\ \dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{29}{24}\\ x=\dfrac{29}{24}:\dfrac{1}{2}\\ x=\dfrac{29}{12}\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\\ \dfrac{1}{2}x=\dfrac{-7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{-13}{24}\\ x=\dfrac{-13}{24}:\dfrac{1}{2}\\ x=\dfrac{-13}{12}\)
i)
\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=0-\dfrac{1}{9}\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}:3\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{27}\\ \left(3x-\dfrac{1}{2}\right)^3=\left(\dfrac{-1}{3}\right)^3\\ \Leftrightarrow3x-\dfrac{1}{2}=\dfrac{-1}{3}\\ 3x=\dfrac{-1}{3}+\dfrac{1}{2}\\ 3x=\dfrac{1}{6}\\ x=\dfrac{1}{6}:3\\ x=\dfrac{1}{18}\)
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