\(\dfrac{9}{14}-\dfrac{1}{2}\)
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C
6 tháng 3 2023
\(\dfrac{3}{5}+\dfrac{1}{2}+\dfrac{8}{15}\\ =\dfrac{3\times6}{5\times6}+\dfrac{1\times15}{2\times15}+\dfrac{8\times2}{15\times2}\\ =\dfrac{18}{30}+\dfrac{15}{30}+\dfrac{16}{30}\\ =\dfrac{49}{30}\\ \dfrac{6}{9}+\dfrac{14}{18}-\dfrac{5}{6}\\ =\dfrac{6\times2}{9\times2}+\dfrac{14}{18}-\dfrac{5\times3}{6\times3}\\ =\dfrac{12}{18}+\dfrac{14}{18}-\dfrac{15}{18}\\ =\dfrac{11}{18}\)
\(\dfrac{9}{20}-\dfrac{3}{5}:\dfrac{4}{1}\\ =\dfrac{9}{20}-\dfrac{3}{5}\times\dfrac{1}{4}\\ =\dfrac{9}{20}-\dfrac{3}{20}\\ =\dfrac{6}{20}\\ =\dfrac{3}{10}\)
\(\dfrac{1}{6}+\dfrac{2}{3}\times\dfrac{8}{9}\\=\dfrac{1}{6}+\dfrac{16}{27}\\ =\dfrac{1\times9}{6\times9}+\dfrac{16\times2}{27\times2}\\ =\dfrac{9}{54}+\dfrac{32}{54}\\ =\dfrac{41}{54}.\)
13 tháng 7 2023
a: =>x-5/14=6/14-1/14=5/14
=>x=10/14=5/7
b; =>2/9:x=3/6+4/6=7/6
=>x=2/9:7/6=2/9*6/7=4/21
c: =>32:x=8
=>x=4
NN
13 tháng 7 2023
`@` `\text {Answer}`
`\downarrow`
`a,`
`x - 5/14 = 3/7 - 1/14`
`x - 5/14 = 5/14`
`=> x = 5/14 + 5/14`
`=> x = 5/7`
Vậy, `x = 5/7`
`b,`
`2/9 \div x = 1/2 + 2/3`
`2/9 \div x = 7/6`
`x = 2/9 \div 7/6`
`x = 4/21`
Vậy, `x = 4/21`
`c,`
\(\dfrac{6}{32\div x}=\dfrac{12}{16}\)
`6/(32 \div x) = 3/4`
`32 \div x = 6 \div 3/4`
`32 \div x = 8`
` x = 32 \div 8`
`x = 4`
Vậy, `x = 4`
1 tháng 3 2022
a: =-8/28-7/28=-15/28
b: \(=\dfrac{-4}{18}+\dfrac{3}{7}\cdot\dfrac{14}{15}=\dfrac{-2}{9}+\dfrac{14}{15}=\dfrac{-10+42}{45}=\dfrac{32}{45}\)
c: \(=\dfrac{-3\cdot5+7\cdot2}{20}\cdot\dfrac{-5}{1}-\dfrac{2}{9}\)
\(=\dfrac{-7}{4}-\dfrac{2}{9}=\dfrac{-63}{36}-\dfrac{8}{36}=-\dfrac{71}{36}\)
8 tháng 3 2022
b: \(\Leftrightarrow\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}=9+\dfrac{5}{7}-\dfrac{5}{7}=9\)
=>x-1/2=27
hay x=55/2
c: =>1/2x-3/4=42/63=2/3
=>1/2x=17/12
hay x=17/6
TT
18 tháng 4 2021
a) \(\dfrac{5}{9}:\left(\dfrac{13}{7}+\dfrac{13}{9}\right)-\dfrac{5}{3}\)(chỗ này mk lười chép lại đề)
=\(\dfrac{5}{9}:\dfrac{208}{63}-\dfrac{5}{3}\)
=\(\dfrac{5}{9}.\dfrac{63}{208}-\dfrac{5}{3}\)
=\(\dfrac{5.63}{9.208}-\dfrac{5}{3}\)
=\(\dfrac{5.7}{1.208}-\dfrac{5}{3}\)
=\(\dfrac{36}{208}-\dfrac{5}{3}\)
=\(\dfrac{108}{624}-\dfrac{1040}{624}\)
=\(\dfrac{-932}{624}\)
=\(\dfrac{233}{156}\)
còn câu b mk chưa học nên mk chịu
18 tháng 4 2021
Giải:
5/9:13/7+5/9:13/9 -1 2/3
=5/9.7/13+5/9.9/13-5/3
=5/9.(7/13+9/13)-5/3
=5/9.16/13-5/3
=80/117-5/3
=-115/117
4 2/5 : 0,5% -1 3/7 .14% +(-0,5)
=22/5:1/200-10/7.7/50 +(-1/2)
=880-1/5-1/2
=8793/10
MH
31 tháng 1 2021
1/ \(\dfrac{4x+7}{x-1}=\dfrac{12x+5}{3x+4}\) (1)
Điều kiện: \(\left\{{}\begin{matrix}x-1\ne0\\3x+4\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-\dfrac{4}{3}\end{matrix}\right.\)
(1) \(\Leftrightarrow\left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)\\\Leftrightarrow12x^2+16x+21x+28=12x^2-12x+5x-5\\ \Leftrightarrow\left(16+21+12-5\right)x=-5-28\\ \Leftrightarrow44x=-33\\ \Leftrightarrow x=-\dfrac{3}{4}\) (Thỏa mãn)
Vậy \(x=-\dfrac{3}{4}\).
2/ \(\dfrac{x}{x-1}-\dfrac{2x}{x^2-1}=0\) (2)
Điều kiện: \(x\ne\pm1\)
(2)\(\Leftrightarrow\dfrac{x}{x-1}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x\left(x+1\right)-2x}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow x\left(x+1\right)-2x=0\\ \Leftrightarrow x^2+x-2x=0\\ \Leftrightarrow x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
So sánh với điều kiện \(\Rightarrow x=0\) là nghiệm của PT.
3/ \(\dfrac{1}{3-x}-\dfrac{14}{x^2-9}=1\) (3)
Điều kiện: \(x\ne\pm3\)
(3)\(\Leftrightarrow\dfrac{1}{3-x}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=1\\ \Leftrightarrow-\dfrac{\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ \Leftrightarrow-\left(x+3\right)-14=\left(x-3\right)\left(x+3\right)\\ \Leftrightarrow-x-17=x^2-9\Leftrightarrow x^2+x+8=0\) (Vô nghiệm do \(x^2+x+8>0\qquad\forall x\)).
Vậy PT vô nghiệm.
4/ \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\) (4)
Điều kiện: \(x\ne\pm1\)
(4)\(\Leftrightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\\ \Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\\ \Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=4\\ \Leftrightarrow\left(x^2+2x+1\right)-\left(x^2-2x+1\right)=4\Leftrightarrow4x=4\Leftrightarrow x=1\) (loại)
Vậy PT vô nghiệm.
5/ \(x+\dfrac{1}{x}=x^2+\dfrac{1}{x^2}\) (5)
Điều kiện: \(x\ne0\)
(5)\(\Leftrightarrow x+\dfrac{1}{x}=\left(x+\dfrac{1}{x}\right)^2-2\)
Đặt \(t=x+\dfrac{1}{x}\), ta có: \(t=t^2-2\\ \Leftrightarrow t^2-t-2=0\Leftrightarrow\left(t-2\right)\left(t+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=2\\t=-1\end{matrix}\right.\)
Với \(t=2\) ta có: \(x+\dfrac{1}{x}=2\Leftrightarrow x^2+1=2x\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\) (thỏa mãn)
Với \(t=-1\) ta có: \(x+\dfrac{1}{x}=-1\Leftrightarrow x^2+1=-x\Leftrightarrow x^2+x+1=0\) (vô nghiệm).
Vậy \(x=1\) là nghiệm PT.
6/ \(\dfrac{x-1}{x^2+4}=\dfrac{x-1}{x+1}\) (6)
Điều kiện: \(x\ne-1\)
(6)\(\Leftrightarrow\dfrac{x-1}{x^2+4}-\dfrac{x-1}{x+1}=0\\ \Leftrightarrow\left(x-1\right)\left(\dfrac{1}{x^2+4}-\dfrac{1}{x+1}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\\dfrac{1}{x^2+4}-\dfrac{1}{x+1}=0\end{matrix}\right.\)
\(x-1=0\Leftrightarrow x=1\) (Thỏa mãn)
\(\dfrac{1}{x^2+4}-\dfrac{1}{x+1}=0\Leftrightarrow\dfrac{1}{x^2+4}=\dfrac{1}{x+1}\Leftrightarrow x^2+4=x+1\\ \Leftrightarrow x^2-x+3=0\) (vô nghiệm).
Vậy \(x=1\) là nghiệm PT.
31 tháng 1 2021
1) ĐKXĐ: \(x\notin\left\{1;-\dfrac{4}{3}\right\}\)
Ta có: \(\dfrac{4x+7}{x-1}=\dfrac{12x+5}{3x+4}\)
\(\Leftrightarrow\left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)\)
\(\Leftrightarrow12x^2+16x+21x+28=12x^2+12x+5x-5\)
\(\Leftrightarrow12x^2+37x+28-12x^2-17x+5=0\)
\(\Leftrightarrow20x+33=0\)
\(\Leftrightarrow20x=-33\)
\(\Leftrightarrow x=-\dfrac{33}{20}\)(nhận)
Vậy: \(S=\left\{-\dfrac{33}{20}\right\}\)
2) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
Ta có: \(\dfrac{x}{x-1}-\dfrac{2x}{x^2-1}=0\)
\(\Leftrightarrow\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}=0\)
Suy ra: \(x^2+x-2x=0\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=1\left(loại\right)\end{matrix}\right.\)
Vậy: S={0}
3) ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
Ta có: \(\dfrac{1}{3-x}-\dfrac{14}{x^2-9}=1\)
\(\Leftrightarrow\dfrac{-1}{x-3}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=1\)
\(\Leftrightarrow\dfrac{-\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(-x-3-14=x^2-9\)
\(\Leftrightarrow x^2-9=-x-17\)
\(\Leftrightarrow x^2-9+x+17=0\)
\(\Leftrightarrow x^2+x+8=0\)
\(\Leftrightarrow x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{31}{4}=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{31}{4}=0\)(vô lý)
Vậy: \(S=\varnothing\)
4) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
Ta có: \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)
Suy ra: \(x^2+2x+1-\left(x^2-2x+1\right)=4\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1=4\)
\(\Leftrightarrow4x=4\)
hay x=1(loại)
Vậy: \(S=\varnothing\)
5) ĐKXĐ: \(x\ne0\)
Ta có: \(x+\dfrac{1}{x}=x^2+\dfrac{1}{x^2}\)
\(\Leftrightarrow\dfrac{x^2+1}{x}=\dfrac{x^4+1}{x^2}\)
\(\Leftrightarrow x^2\left(x^2+1\right)=x\left(x^4+1\right)\)
\(\Leftrightarrow x^4+x^2=x^5+x\)
\(\Leftrightarrow x^5+x-x^4-x^2=0\)
\(\Leftrightarrow x\left(x^4-x^3-x+1\right)=0\)
\(\Leftrightarrow x\left[x^3\left(x-1\right)-\left(x-1\right)\right]=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x^3-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)^2\cdot\left(x^2+x+1\right)=0\)
mà \(x^2+x+1>0\)
nên \(x\cdot\left(x-1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x-1=0\end{matrix}\right.\Leftrightarrow x=1\)
Vậy: S={1}
6) ĐKXĐ: \(x\in R\)
Ta có: \(\dfrac{x-1}{x^2+4}=\dfrac{x-1}{x+1}\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left(x^2+4\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)-\left(x-1\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1-x^2-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-x^2+x-3\right)=0\)
\(\Leftrightarrow-\left(x-1\right)\left(x^2-x+3\right)=0\)
mà \(x^2-x+3>0\)
nên x-1=0
hay x=1(nhận)
Vậy: S={1}
28 tháng 1
\(\dfrac{14}{7}>\dfrac{9}{7}\)
\(\dfrac{6}{13}< 1\)
\(\dfrac{7}{9}=\dfrac{14}{18}\)
\(\dfrac{4}{9}=\dfrac{20}{45}< \dfrac{26}{45}\)
\(\dfrac{11}{24}>\dfrac{9}{24}=\dfrac{3}{8}\)
\(\dfrac{42}{91}=\dfrac{42:7}{91:7}=\dfrac{6}{13}< \dfrac{7}{13}\)
\(\dfrac{34}{64}< \dfrac{52}{64}=\dfrac{13}{16}\)
\(\dfrac{17}{30}>\dfrac{16}{30}=\dfrac{8}{15}\)
17 tháng 7 2021
a) Ta có: \(\dfrac{2}{3}x-1=\dfrac{3}{2}\)
\(\Leftrightarrow x\cdot\dfrac{2}{3}=\dfrac{5}{2}\)
hay \(x=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{5}{2}\cdot\dfrac{3}{2}=\dfrac{15}{4}\)
b) Ta có: \(\left|5x-\dfrac{1}{2}\right|-\dfrac{2}{7}=25\%\)
\(\Leftrightarrow\left|5x-\dfrac{1}{2}\right|=\dfrac{1}{4}+\dfrac{2}{7}=\dfrac{15}{28}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{1}{2}=\dfrac{15}{28}\\5x-\dfrac{1}{2}=\dfrac{-15}{28}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{29}{28}\\5x=\dfrac{-1}{28}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{140}\\x=\dfrac{-1}{140}\end{matrix}\right.\)
17 tháng 7 2021
c) Ta có: \(\dfrac{x-3}{4}=\dfrac{16}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=64\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=8\\x-3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\)
d) Ta có: \(\dfrac{-8}{13}+\dfrac{7}{17}+\dfrac{21}{31}\le x\le\dfrac{-9}{14}+4-\dfrac{5}{14}\)
\(\Leftrightarrow\dfrac{3246}{6851}\le x\le3\)
\(\Leftrightarrow x\in\left\{1;2;3\right\}\)
9/14 - 1/2 = 9/14 - 7/14 = 2/14 = 1/7