a, \(\dfrac{3}{7}+\dfrac{4}{7}x=\dfrac{1}{3}\)

\(\Rightarrow\) \(\dfrac{4}{7}x=\dfrac{1}{3}-\dfrac{3}{7}\)

\(\Rightarrow\) \(\dfrac{4}{7}x=\dfrac{-2}{21}\)

\(\Rightarrow x=\dfrac{-2}{21}:\dfrac{4}{7}\)

\(\Rightarrow x=\dfrac{-1}{6}\)

b, \(25-\left(5-x\right)=-7\)

\(\Rightarrow\) \(5-x=25-\left(-7\right)\)

\(\Rightarrow5-x=32\)

\(\Rightarrow x=5-32\)

\(\Rightarrow x=-27\)

c, \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Rightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)

\(\Rightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)

\(\Rightarrow x=\dfrac{1}{4}:\dfrac{-7}{20}\)

\(\Rightarrow x=\dfrac{-5}{7}\)

d, \(2x\left(x-\dfrac{1}{7}\right)=0\)

\(\Rightarrow\) \(\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0:2\\x=0+\dfrac{1}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)

e, \(\left|\dfrac{1}{2}x-\dfrac{3}{4}\right|-7=-3\)

\(\Rightarrow\left|\dfrac{1}{2}x-\dfrac{3}{4}\right|=-3+7\)

\(\Rightarrow\left|\dfrac{1}{2}x-\dfrac{3}{4}\right|=4\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{3}{4}=4\\\dfrac{1}{2}x-\dfrac{3}{4}=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=4+\dfrac{3}{4}\\\dfrac{1}{2}x=-4+\dfrac{3}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{19}{4}\\\dfrac{1}{2}x=\dfrac{-13}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19}{4}:\dfrac{1}{2}\\x=\dfrac{-13}{4}:\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19}{2}\\x=\dfrac{-13}{2}\end{matrix}\right.\)

a)\(\dfrac{3}{7}+\dfrac{4}{7}x=\dfrac{1}{3}\)

\(\dfrac{4}{7}x=\dfrac{1}{3}-\dfrac{3}{7}\)

\(\dfrac{4}{7}x=\dfrac{-2}{21}\)

\(x=\dfrac{-2}{21}:\dfrac{4}{7}\)

\(x=\dfrac{-1}{6}\)

b)\(25-\left(5-x\right)=-7\)

\(5-x=25-\left(-7\right)\)

\(5-x=32\)

x= -27

c)\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)

\(\dfrac{1}{4}:x=\dfrac{-7}{20}\)

\(x=\dfrac{1}{4}:\dfrac{-7}{20}\)

\(x=\dfrac{-5}{7}\)

d)\(2x\left(x-\dfrac{1}{7}\right)=0\)

⇒\(\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)

e)\(|\dfrac{1}{2}x-\dfrac{3}{7}|-7=-3\)

\(\left|\dfrac{1}{2}x-\dfrac{3}{7}\right|=-3+7\)

\(\left|\dfrac{1}{2}x-\dfrac{3}{7}\right|=4\)

⇒\(\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{3}{4}=4\\\dfrac{1}{2}x-\dfrac{3}{4}=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=4\dfrac{3}{4}\Rightarrow x=9\dfrac{1}{2}=\dfrac{19}{2}\\\dfrac{1}{2}x=-3\dfrac{1}{4}\Rightarrow x=\dfrac{-13}{2}\end{matrix}\right.\)

Bài 1:

a) Ta có: \(\frac{-5}{8}+x=\frac{4}{9}\)

\(\Leftrightarrow x=\frac{4}{9}-\frac{-5}{8}=\frac{32}{72}-\frac{-45}{72}\)

hay \(x=\frac{77}{72}\)

Vậy: \(x=\frac{77}{72}\)

b) Ta có: \(1\frac{3}{4}\cdot x+1\frac{1}{2}=-\frac{4}{5}\)

\(\Leftrightarrow\frac{7}{4}\cdot x+\frac{3}{2}=-\frac{4}{5}\)

\(\Leftrightarrow\frac{7}{4}\cdot x=-\frac{4}{5}-\frac{3}{2}=-\frac{23}{10}\)

\(\Leftrightarrow x=\frac{-23}{10}:\frac{7}{4}=\frac{-23}{10}\cdot\frac{4}{7}\)

hay \(x=-\frac{46}{35}\)

Vậy: \(x=-\frac{46}{35}\)

c) Ta có: \(\frac{1}{4}+\frac{3}{4}x=\frac{3}{4}\)

\(\Leftrightarrow\frac{3}{4}x=\frac{2}{4}\)

\(\Leftrightarrow x=\frac{2}{4}:\frac{3}{4}=\frac{2}{4}\cdot\frac{4}{3}\)

hay \(x=\frac{2}{3}\)

Vậy: \(x=\frac{2}{3}\)

d) Ta có: \(x\cdot\left(\frac{1}{4}+\frac{1}{5}\right)-\left(\frac{1}{7}+\frac{1}{8}\right)=0\)

\(\Leftrightarrow x\cdot\frac{9}{20}-\frac{15}{56}=0\)

\(\Leftrightarrow x\cdot\frac{9}{20}=\frac{15}{56}\)

\(\Leftrightarrow x=\frac{15}{56}:\frac{9}{20}=\frac{15}{56}\cdot\frac{20}{9}\)

hay \(x=\frac{25}{42}\)

Vậy: \(x=\frac{25}{42}\)

e) Ta có: \(\frac{3}{35}-\left(\frac{3}{5}+x\right)=\frac{2}{7}\)

\(\Leftrightarrow\frac{3}{35}-\frac{3}{5}-x=\frac{2}{7}\)

\(\Leftrightarrow\frac{-18}{35}-x=\frac{2}{7}\)

\(\Leftrightarrow-x=\frac{2}{7}-\frac{-18}{35}=\frac{2}{7}+\frac{18}{35}=\frac{4}{5}\)

hay \(x=-\frac{4}{5}\)

Vậy: \(x=-\frac{4}{5}\)

f) Ta có: \(\frac{3}{7}+\frac{1}{7}:x=\frac{3}{14}\)

\(\Leftrightarrow\frac{1}{7}\cdot\frac{1}{x}=\frac{3}{14}-\frac{3}{7}=\frac{-3}{14}\)

\(\Leftrightarrow\frac{1}{x}=\frac{-3}{14}:\frac{1}{7}=-\frac{3}{14}\cdot7=-\frac{3}{2}\)

\(\Leftrightarrow x=\frac{1\cdot2}{-3}=\frac{2}{-3}=-\frac{2}{3}\)

Vậy: \(x=-\frac{2}{3}\)

g) Ta có: \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\2x-\frac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{5}\\x=\frac{1}{3}:2=\frac{1}{6}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{1}{5};\frac{1}{6}\right\}\)

Ta có \(5^5-5^4+5^3=5^3\left(5^2-5+1\right)=5^3.21=5^3.3.7\)

Vì 5³.3 là số nguyên nên \(5^3.3.7⋮7\)

Vậy  \(5^5-5^4+5^3⋮7\)

c) \(3^{x+3}+3^{x+1}+2^{x+3}+2^{x+2}\)

\(=\left(3^{x+3}+3^{x+1}\right)+\left(2^{x+3}+2^{x+2}\right)\)

\(=3^x\left(3^2+3\right)+2^x\left(2^2+2\right)\)

\(=3^x.12+2^x.6\)

\(=6\left(2.3^x+2^x\right)\)

Vì \(2.3^x+2^x\in Z\)

Nên : \(6\left(2.3^x+2^x\right)⋮6\)

Vậy \(3^{x+3}+3^{x+1}+2^{x+3}+2^{x+2}⋮6\)

a) ta có : \(5^5-5^4+5^3=5^3.\left(5^2-5+1\right)=5^3.\left(25-5+1\right)\)

\(5^3.21=5^3.3.7⋮7\) (đpcm)

b) ta có : \(7^6+7^5-7^4=7^4.\left(7^2+7-1\right)=7^4.\left(49+7-1\right)\)

\(=7^4.55=7^4.5.11⋮11\) (đpcm)

c) ta có : \(3^{x+2}-2^{x+3}+3^x-2^{x+1}=3^{x+2}+3^x-2^{x+3}-2^{x+1}\)

\(=3^x\left(3^2+1\right)-2^x\left(2^3+2\right)=3^x.\left(9+1\right)-2^x.\left(8+2\right)\)

\(=3^x.10-2^x.10=10\left(3^x-2^x\right)⋮10\) (đpcm)

d) \(3^{x+3}+3^{x+1}+2^{x+3}+2^{x+2}=3^x.\left(3^3+3\right)+2^x.\left(2^3+2^2\right)\)

\(=3^x.\left(27+3\right)+2^x\left(8+4\right)=3^x.30+2^x.12=6.\left(3^x.5+2^x.2\right)⋮6\) (đpcm)

a)Ta có:\(5^5-5^4+5^3=5^3\left(5^2-5+1\right)=5^3.21\)(vì 21 chia hết cho 7)

\(\)\(\RightarrowĐPCM\)

b)Ta có: \(7^6+7^5-7^4⋮11=7^4\left(7^2+7-1\right)=7^4.55⋮11\)

\(\Rightarrowđpcm\)

1: \(\Leftrightarrow\left(x+1\right)^2=4\)

=>x+1=2 hoặc x+1=-2

=>x=1 hoặc x=-3

2: \(\Leftrightarrow7x-21=5x+25\)

=>2x=46

=>x=23

3: \(\Leftrightarrow x^2+4x+3=x^2+0.5x+4x+2\)

=>4,5x+2=4x+3

=>x=1

2, \(\Rightarrow\left\{{}\begin{matrix}\\\dfrac{5}{4}x-\dfrac{7}{2}=0\\\dfrac{5}{8}x+\dfrac{3}{5}=0\\\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{14}{5}\\\\x=\dfrac{-24}{25}\\\end{matrix}\right.\)

7/4.x+3/2=-4/5

7/4.x=-4/5-3/2

7/4.x=-23/10

x=-23/10:7/4

x=-46/35

vậy x=-46/35

1/4+3/4.x=3/4

1.x=3/4

x=3/4:1

x=3/4

vậy x=3/4

x.(1/4+1/5)-(1/7+1/8)=0

x.9/20-15/56=0

x.51/280=0

x=0:51/280

x=0

vậy x=0

3/35-(3/5+x)=2/7

(3/5+x)=3/35-2/7

(3/35+x)=-1/5

x=-1/5-3/5

x=-4/5

vậy x=-4/5

\(a,1\frac{3}{4}.x+1\frac{1}{2}=\frac{4}{5}\)

\(\frac{7}{4}.x=\frac{4}{5}-\frac{3}{2}\)

\(\frac{7}{4}.x=\frac{-7}{10}\)

\(x=\frac{-7}{10}:\frac{7}{4}\)

\(x=\frac{-2}{5}\)

\(b,\frac{1}{4}+\frac{3}{4}.x=\frac{3}{4}\)

\(\frac{3}{4}.x=\frac{3}{4}-\frac{1}{4}\)

\(\frac{3}{4}.x=\frac{1}{2}\)

\(x=\frac{1}{2}:\frac{3}{4}\)

\(x=\frac{2}{3}\)

\(c,x.\left(\frac{1}{4}+\frac{1}{5}\right)-\left(\frac{1}{7}+\frac{1}{8}\right)=0\)

\(x.\frac{9}{20}-\frac{15}{56}=0\)

\(x.\frac{9}{20}=\frac{15}{56}\)

\(x=\frac{15}{56}:\frac{9}{20}\)

\(x=\frac{25}{42}\)

\(d,\frac{3}{35}-\left(\frac{3}{5}+x\right)=\frac{2}{7}\)

\(\frac{3}{5}+x=\frac{3}{35}-\frac{2}{7}\)

\(\frac{3}{5}+x=\frac{-1}{5}\)

\(x=\frac{-1}{5}-\frac{3}{5}\)

\(x=\frac{-4}{5}\)

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