a.

31,5 – x = (18,6 – 12,3) : 3

31,5 - x = 2,1

x = 31,5 - 2,1

x = 29,4

b.

???

Bài 2. Tìm x biết:

a) 31,5 – x = (18,6 – 12,3) : 3

31,5 – x = 6,3 : 3

31,5 – x = 2,1

x = 31,5 – 2,1

x = 29,4

b) Đề sai.

a, X = 29,4

a: \(x+\dfrac{3}{9}=\dfrac{7}{6}\cdot\dfrac{2}{3}\)

=>\(x+\dfrac{1}{3}=\dfrac{14}{18}=\dfrac{7}{9}\)

=>\(x=\dfrac{7}{9}-\dfrac{1}{3}=\dfrac{7}{9}-\dfrac{3}{9}=\dfrac{4}{9}\)

b: \(x-\dfrac{2}{3}=\dfrac{1}{8}:\dfrac{5}{4}\)

=>\(x-\dfrac{2}{3}=\dfrac{1}{8}\cdot\dfrac{4}{5}=\dfrac{1}{10}\)

=>\(x=\dfrac{1}{10}+\dfrac{2}{3}=\dfrac{3+20}{30}=\dfrac{23}{30}\)

TThế giới oi oi oi 

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

=> 3/5 + x= 3/35- 2/7

=> 3/5 +x = -1/5

=> x = -1/5 -3/5

=> x = -4/5

b) 3/7 +1/7 : x = 3/14

=> 1/7 : x= 3/14 -3/7

=> 1/7 : x = -3/14

=> x = 1/7 : -3/14 

=> x = -2/3

c) (5x-1).(2x-1/3)=0

=> \(\left[{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}5x=0+1=1\\2x=0+\dfrac{1}{3}=\dfrac{1}{3}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{3}:2=\dfrac{1}{6}\end{matrix}\right.\)

Học tốt :D

a)x=-4/5

b)x=-2/3

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

Vậy.........

mik lười mong bn thông cảm

a) \(\dfrac{3}{35}-\left(\dfrac{3}{5}+x\right)=\dfrac{2}{7}\\ \Rightarrow\dfrac{3}{5}+x=-\dfrac{1}{5}\\ \Rightarrow x=-\dfrac{4}{5}\)

b) \(\dfrac{3}{7}+\dfrac{1}{7}:x=\dfrac{3}{14}\\ \Rightarrow\dfrac{1}{7}:x=-\dfrac{3}{14}\\ \Rightarrow x=-\dfrac{2}{3}\)

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

a) \(x+1^3=2^5-\left(-1^3\right)\)

\(\Rightarrow x+1=33\)

=> x = 32

b) \(3^7-x=1^4-\left(-3^5\right)\)

\(\Rightarrow2187-x=1+243=244\)

=> x = 1943

a) \(\Leftrightarrow x+1=32+1\)

\(\Leftrightarrow x=32\)

Vậy x = 32

b) \(\Leftrightarrow2187-x=1+243\)

\(\Leftrightarrow2187-x=244\)

\(\Leftrightarrow x=1943\)

Vậy x = 1943

a) \(x+1^3=2^5-\left(-1^3\right)\) 

      \(x+1=32-\left(-1\right)\) 

      \(x+1=33\) 

            \(x=33-1\) 

            \(x=32\) 

b) \(3^7-x=1^4-\left(-3^5\right)\) 

\(2187-x=1-\left(-243\right)\) 

\(2187-x=244\) 

            \(x=2187-244\) 

            \(x=1943\)

1)

\(x-17=23\\ \Rightarrow x=23+17\\ \Rightarrow x=40\)

2)

\(2\left(x-1\right)=7+\left(-3\right)\\ \Rightarrow2x-2=4\\ \Rightarrow2x=4+2\\ \Rightarrow2x=8\\ \Rightarrow x=4\)

3)

\(4\left(x+5\right)^3-7=101\\ \Rightarrow4\left(x+5\right)^3=101+7\\ \Rightarrow4\left(x+5\right)^3=108\\ \Rightarrow\left(x+5\right)^3=108\div4\\ \Rightarrow\left(x+5\right)^3=27\\ \Rightarrow\left(x+5\right)^3=3^3\\ \Rightarrow x+5=3\Rightarrow x=3-5\\ \Rightarrow x=-2\)

4)

\(2^{x+1}\times3+15=39\\ \Rightarrow2^{x+1}\times3=39-15\\ \Rightarrow2^{x+1}\times3=24\\ \Rightarrow2^{x+1}=24\div3\\ \Rightarrow2^{x+1}=8\)

\( \Rightarrow2^{2+1}=8\\\Rightarrow2^3=8\Rightarrow x=2 \)

a) x - 17 = 23

x = 23 + 17

x = 40

Vậy x = 40

b) 2 ( x - 1 ) = 7 + ( - 3 )

x - 1 = 4 : 2

x = 2 + 1

x = 3

Vậy x = 3

c) 4 ( x + 3 )^3 - 7 = 101

4 ( x + 3 )^3 = ( 101 + 7 ) : 4

( x + 3 )^3 = 3^3

⇒ x + 3 = 3

⇒ x = 0

Vậy x = 0

d) 2^{ x+ 1 } . 3 + 15 = 39

2^{ x + 1 } = ( 39 - 15 ) : 3

2^{ x + 1 } = 2{ 2 + 1 }

⇒ x + 1 = 2 + 1

⇒ x = 2

Vậy x = 2

a: \(x\cdot\dfrac{2}{5}+\dfrac{1}{2}\cdot x=9\)

=>\(x\left(\dfrac{2}{5}+\dfrac{1}{2}\right)=9\)

=>\(x\cdot\dfrac{9}{10}=9\)

=>\(x=9:\dfrac{9}{10}=10\)

b: \(\dfrac{1}{9}:x+\dfrac{3}{9}:x=\dfrac{5}{7}\)

=>\(\left(\dfrac{1}{9}+\dfrac{3}{9}\right):x=\dfrac{5}{7}\)

=>\(\dfrac{4}{9}:x=\dfrac{5}{7}\)

=>\(x=\dfrac{4}{9}:\dfrac{5}{7}=\dfrac{4}{9}\cdot\dfrac{7}{5}=\dfrac{28}{45}\)

2:

a: Gọi d=ƯCLN(4n+7;2n+3)

=>\(\left\{{}\begin{matrix}4n+7⋮d\\2n+3⋮d\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4n+7⋮d\\4n+6⋮d\end{matrix}\right.\Leftrightarrow1⋮d\)

=>d=1

=>ƯCLN(4n+7;2n+3)=1

b: Gọi \(d=ƯCLN\left(3n+5;6n+9\right)\)

=>\(\left\{{}\begin{matrix}3n+5⋮d\\6n+9⋮d\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6n+10⋮d\\6n+9⋮d\end{matrix}\right.\)

=>\(1⋮d\)

=>d=1

=>Đây là phân số tối giản

\((2x-1)^2+(x+3)^2-5(x+7)(x-7)=0\)

\(< =>4x^2-4x+1+x^2+6x+9-5\left(x^2-7^2\right)=0\\ < =>4x^2-4x+1+x^2+6x+9-5x^2+245=0\\ < =>2x+255=0\\ < =>2x=-255=>x=\dfrac{-255}{2}\)

Vậy \(x=\dfrac{-255}{2}\)

\(\Rightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)

\(\Rightarrow2x+255=0\Rightarrow2x=-255\Rightarrow x=-\dfrac{255}{2}\)

Lời giải:

a) 

$\frac{4}{7}x=\frac{2}{3}+\frac{1}{5}=\frac{13}{15}$

$x=\frac{13}{15}:\frac{4}{7}=\frac{91}{60}$

b) 

$\frac{5}{7}:x=\frac{1}{6}-\frac{4}{5}$

$\frac{5}{7}:x=\frac{-19}{30}$

$x=\frac{5}{7}:\frac{-19}{30}=\frac{-150}{133}$

a) \(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{1}{5}\) 

            \(\dfrac{4}{7}.x=\dfrac{1}{5}+\dfrac{2}{3}\) 

            \(\dfrac{4}{7}.x=\dfrac{13}{15}\) 

                \(x=\dfrac{13}{15}:\dfrac{4}{7}\) 

                \(x=\dfrac{91}{60}\) 

b) \(\dfrac{4}{5}+\dfrac{5}{7}:x=\dfrac{1}{6}\) 

            \(\dfrac{5}{7}:x=\dfrac{1}{6}-\dfrac{4}{5}\) 

            \(\dfrac{5}{7}:x=\dfrac{-19}{30}\) 

                  \(x=\dfrac{5}{7}:\dfrac{-19}{30}\) 

                 \(x=\dfrac{-150}{133}\)

\(a,\Leftrightarrow4x^2-20x-4x^2+7x-3=23\\ \Leftrightarrow-13x=-26\\ \Leftrightarrow x=2\\ b,\Leftrightarrow x^2+4x+4+4x^2-12x+9=5x^2+35x\\ \Leftrightarrow-43x=-13\\ \Leftrightarrow x=\dfrac{13}{43}\)

a) \(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=23\)

\(\Leftrightarrow4x^2-20x-4x^2+7x-3=23\)

\(\Leftrightarrow13x=-26\Leftrightarrow x=-2\)

b) \(\left(x+2\right)^2+\left(2x-3\right)^2=5x\left(x+7\right)\)

\(\Leftrightarrow x^2+4x+4+4x^2-12x+9=5x^2+35x\)

\(\Leftrightarrow43x=13\Leftrightarrow x=\dfrac{13}{43}\)