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`@` `\text {Ans}`
`\downarrow`
`1)`
\(x+\dfrac{1}{2}=\dfrac{5}{3}\)
`\Rightarrow` \(x=\dfrac{5}{3}-\dfrac{1}{2}\)
`\Rightarrow`\(x=\dfrac{7}{6}\)
Vậy, `x =`\(\dfrac{7}{6}\)
`2)`
\(\dfrac{3}{5}-x=\dfrac{1}{3}\)
`\Rightarrow`\(x=\dfrac{3}{5}-\dfrac{1}{3}\)
`\Rightarrow`\(x=\dfrac{4}{15}\)
Vậy, `x =`\(\dfrac{4}{15}\)
`3)`
\(\dfrac{3}{4}+x=\dfrac{7}{2}\)
`\Rightarrow`\(x=\dfrac{7}{2}-\dfrac{3}{4}\)
`\Rightarrow`\(x=\dfrac{11}{4}\)
Vậy, \(x=\dfrac{11}{4}\)
`4)`
\(x-\dfrac{4}{3}=\dfrac{7}{9}\)
`\Rightarrow`\(x=\dfrac{7}{9}+\dfrac{4}{3}\)
`\Rightarrow`\(x=\dfrac{19}{9}\)
Vậy, `x=`\(\dfrac{19}{9}\)
`5)`
\(x-\dfrac{5}{6}=\dfrac{7}{3}\)
`\Rightarrow`\(x=\dfrac{7}{3}+\dfrac{5}{6}\)
`\Rightarrow x =`\(\dfrac{19}{6}\)
Vậy, `x=`\(\dfrac{19}{6}\)
`6)`
\(x-\dfrac{1}{5}=\dfrac{9}{10}\)
`\Rightarrow x=`\(\dfrac{9}{10}+\dfrac{1}{5}\)
`\Rightarrow x=`\(\dfrac{11}{10}\)
Vậy, `x=`\(\dfrac{11}{10}\)
\(1,\\ x+\dfrac{1}{2}=-\dfrac{5}{3}\\ x=-\dfrac{5}{3}-\dfrac{1}{2}\\ x=-\dfrac{13}{6}\\ Vậyx=-\dfrac{13}{6}\)
\(2,\\ \dfrac{1}{3}-x=\dfrac{3}{5}\\ x=\dfrac{1}{3}-\dfrac{3}{5}\\ x=-\dfrac{4}{15}\\ Vậyx=-\dfrac{4}{15}\)
\(3,\\ 3-4+x=\dfrac{7}{2}\\ -1+x=\dfrac{7}{2}\\ x=\dfrac{7}{2}+1\\ x=\dfrac{9}{2}\\ Vậyx=\dfrac{9}{2}\)
\(4,\\ x-\dfrac{4}{3}=-\dfrac{7}{9}\\ x=-\dfrac{7}{9}+\dfrac{4}{3}\\ x=\dfrac{15}{27}\\ Vậyx=\dfrac{15}{27}\)
\(5,\\ x-\left(-\dfrac{7}{3}\right)=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{3}\\ x=-\dfrac{27}{18}\\ Vậyx=-\dfrac{27}{18}\)
\(6,\\ x-\dfrac{1}{5}=\dfrac{9}{10}\\ x=\dfrac{9}{10}+\dfrac{1}{5}\\ x=\dfrac{11}{10}\\ Vậyx=\dfrac{11}{10}\)
\(7,\\ x+\dfrac{5}{12}=\dfrac{3}{8}\\ x=\dfrac{3}{8}-\dfrac{5}{12}\\ x=-\dfrac{1}{24}\\ Vậyx=-\dfrac{1}{24}\)
\(8,\\ x+\dfrac{5}{4}=\dfrac{7}{6}\\ x=\dfrac{7}{6}-\dfrac{5}{4}\\ x=-\dfrac{9}{24}\\ Vậyx=-\dfrac{9}{24}\)
\(9,\\ x-\dfrac{2}{7}=\dfrac{1}{35}\\ x=\dfrac{1}{35}+\dfrac{2}{7}\\ x=\dfrac{11}{35}\\ Vậyx=\dfrac{11}{35}\\ 10,\\ x-\dfrac{1}{5}=-\dfrac{7}{10}\\ x=-\dfrac{7}{10}+\dfrac{1}{5}\\ x=-\dfrac{1}{2}\\ Vậyx=-\dfrac{1}{2}\)
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1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}.3\)
\(\left|4-2x\right|=1\)
=>\(4-2x=\pm1\)
+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)
\(2x=4-1\) \(2x=4-\left(-1\right)\)
\(2x=3\) \(2x=4+1\)
\(x=3:2\) \(2x=5\)
\(x=1,5\) \(x=5:2\)
Vậy x=1,5 \(x=2,5\)
Vậy x=2,5
2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)
\(9:\left|x+\left(-1\right)\right|=-3\)
\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)
\(\left|x+\left(-1\right)\right|=-3\)
=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )
Vậy x = \(\varnothing\)
b: \(x+\dfrac{5}{6}=\dfrac{3}{8}\)
=>\(x=\dfrac{3}{8}-\dfrac{5}{6}\)
=>\(x=\dfrac{9}{24}-\dfrac{20}{24}=-\dfrac{11}{24}\)
c: \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{5}{6}\)
=>\(\dfrac{2}{3}x=\dfrac{5}{6}+\dfrac{1}{2}=\dfrac{8}{6}=\dfrac{4}{3}\)
=>2x=4
=>x=4/2=2
d: \(\dfrac{x}{7}=\dfrac{6}{-21}\)
=>\(\dfrac{x}{7}=\dfrac{-2}{7}\)
=>x=-2
e: \(\left(\dfrac{7}{3}x-0,6\right):\dfrac{3^2}{5}=1\)
=>\(\dfrac{7}{3}x-0,6=\dfrac{3^2}{5}=1,8\)
=>\(\dfrac{7}{3}x=2,4\)
=>\(x=2,4:\dfrac{7}{3}=2.4\cdot\dfrac{3}{7}=\dfrac{7.2}{7}=\dfrac{36}{35}\)
f: \(\dfrac{x}{45}=\dfrac{5}{6}+\dfrac{-29}{30}\)
=>\(\dfrac{x}{45}=\dfrac{25}{30}-\dfrac{29}{30}=-\dfrac{4}{30}=-\dfrac{2}{15}\)
=>\(x=-\dfrac{2}{15}\cdot45=-6\)
g: \(\left(4,5-2x\right)\cdot\left(-\dfrac{1^4}{7}\right)=\dfrac{11}{14}\)
=>\(4,5-2x=\dfrac{11}{14}:\dfrac{-1}{7}=\dfrac{-11}{2}\)
=>\(2x=4,5+\dfrac{11}{2}=\dfrac{20}{2}=10\)
=>x=10/2=5
h: \(-\dfrac{2}{7}+\dfrac{4}{7}x=\dfrac{5}{7}\)
=>\(\dfrac{4}{7}x=\dfrac{5}{7}+\dfrac{2}{7}=\dfrac{7}{7}\)
=>4x=7
=>\(x=\dfrac{7}{4}\)
a) 3 3/4 . x = 1 1/2
<=> 15/4 . x = 3/2
<=> x = 3/4 . 4/15
<=> x = 1/5
Vậy x = 1/5
b) 1 1/4 x + 1 1/2 = 1 1/4
<=> 5/4 . x + 3/2 = 5/4
<=> 5/4 . x = 5/4 - 3/2
<=> 5/4 . x = -1/4
<=> x = -1/4 . 4/5
<=> x = -1/5
Vậy x = -1/5
c) ( 3 1/3 - 1 1/2 x ) : 5/6 = 1 1/2
<=> ( 10/3 - 3/2 x ) : 5/6 = 3/2
<=> 10/3 - 3/2 x = 3/2 . 5/6
<=> 10/3 - 3/2 x = 5/4
<=> 3/2 . x = 10/3 - 5/4
<=> 3/2 . x = 25/12
<=> x = 25/12 . 2/3
<=> x = 25/18
Vậy x = 25/18
d) ( 3/7 x - 1 ) : 4 = -1/28
<=> 3/7 . x - 1 = -1/28 . 1/4
<=> 3/7 . x - 1 = -1/112
<=> 3/7 . x = -1/112 + 1
<=> 3/7 . x = 111/112
<=> x = 111/112 . 7/3
<=> x = 37/16
Vậy x = 37/16
e) | x - 3/4 | = 1
<=> x - 3/4 = 1
hoặc x - 3/4 = -1
<=> x = 1 + 3/4
hoặc x = -1 + 3/4
<=> x = 7/4
hoặc x = -1/4
Vậy x = 7/4 ; x = -1/4
f) | 2/3 . x + 1/3 | = 5/6
<=> 2/3 . x + 1/3 = 5/6
hoặc 2/3 . x + 1/3 = -5/6
<=> 2/3 . x = 5/6 - 1/3
hoặc 2/3 . x = -5/6 - 1/3
<=> 2/3 . x = 1/2
hoặc 2/3 . x = -7/6
<=> x = 1/2 . 3/2
hoặc x = -7/6 . 3/2
<=> x = 3/4
hoặc x = -7/4
Vậy x = 3/4 ; x = -7/4
1: 2*2*2*2*2*2=2^6
2: x*x*x*x=x^4
3: =7^4*6^4=42^4
4: =5^2*3^2*4^2=60^2
5: =2*3^2*5^5
6: =10^3*10^3=10^6
MẤY DÒNG NÀO BẠN THẤY KO CẦN THIẾT THÌ LƯỢC BỎ NHA!!!
a) \(2\left(x-5\right)-3\left(x+6\right)=4\left(x-7\right)\)
\(2x-10-3x-18=4x-28\)
\(2x-3x-4x-10-18=-28\)
\(-5x-28=-28\)
\(-5x=-28+28=0\)
\(x=\frac{0}{-5}=0\)
b) \(3\left(x-1\right)-2\left(x+5\right)=2\left(x-3\right)\)
\(3x-3-2x-10=2x-6\)
\(3x-2x-2x-3-10=-6\)
\(-x-13=-6\)
\(-x=-6+13=7\)
\(x=-7\)
c) \(5\left(1-x\right)-6\left(1+x\right)=7\left(3-x\right)\)
\(5-5x-6-6x=21-7x\)
\(-5x-6x+7x+5-6=21\)
\(-4x-1=21\)
\(-4x=22\)
\(x=\frac{22}{-4}=\frac{-11}{2}\)
d) \(2x+5-3\left(3x+7\right)=6\left(1-x\right)+8\)
\(2x+5-9x-21=6-6x+8\)
\(2x-9x+6x+5-21=6+8\)
\(-x-16=14\)
\(-x=14+16=30\)
\(x=-30\)
e) \(x-2+3\left(x-4\right)=5\left(x-6\right)+7\)
\(x-2+3x-12=5x-30+7\)
\(x+3x-5x-2-12=-30+7\)
\(-x-14=-23\)
\(-x=-23+14=-9\)
\(x=9\)
f) \(x+2+3\left(1-x\right)-5\left(2-x\right)=6\left(1-x\right)+\left(3-x\right)\)
\(x+2+3-3x-10+5x=6-6x+3-x\)
\(x-3x+5x+6x+x+2+3-10=6+3\)
\(10x-7=9\)
\(10x=9+7=16\)
\(x=\frac{16}{10}=\frac{8}{5}\)
\(\dfrac{4}{6}-\dfrac{2}{3}:x=\dfrac{1}{5}\\ \Rightarrow\dfrac{2}{3}:x=\dfrac{4}{6}-\dfrac{1}{5}\\ \Rightarrow\dfrac{2}{3}:x=\dfrac{7}{15}\\ \Rightarrow x=\dfrac{2}{3}:\dfrac{7}{15}\\ \Rightarrow x=\dfrac{2}{3}\times\dfrac{15}{7}\\ \Rightarrow x=\dfrac{10}{7}\)
Vậy `x=10/7`
ta có 4/6-2/3:x=1/5
=>2/3 : x = 4/6 -1/5
=>2/3 : x = 7/15
=> x =2/3 : 7/15
=>x = 10/7