A) => X = ( 357 - 7 ) : 5

=> X = 350 : 5

 => X = 70

B) 

=> ( x + 5 ) = ( 357 - 7 ) : 5

=> x + 5      = 350 : 5

=> x + 5      = 70

 =>      x       = 70 - 5

=>       x        = 65

357:[x+5]=5 dư 7

\(357\div x=5\) dư 7

\(\left(357-7\right)\div x=5\)

\(350\div x=5\)

\(x=350\div5\)

\(x=70\)

x= (357-7):5= 350:5=70

357 : x = 5 dư 7

357 - 7 = x × 5

350 = x × 5

x = 350 : 5

x = 70

357 : X =5(dư 7)

        X  = ( 357 - 7 ) : 5

       X   = 350 : 5

       X   = 70

Vậy X = 70.

X : 23 =45(dư 6)

X        = 45 x 23 + 6

X        = 1041

Vậy X = 1041.

7 x (X-11)-6=757

7 x (X-11)   = 757 + 6

7 x (X-11)    = 763

      X - 11    = 763 : 7

      X - 11    = 109

     X            = 109 + 11

      X           = 120

Vậy X = 120.

Hok tốt !

x bawngf120 hok tốt 

nha

357:x=5 dư 7

(357-7):x=5

350:x=5

      x=350:5

      x=70

357:70=5 dư 7

357:x =5 dư 7

357 : x = 12

x =      = 357 : 12 

x          = 29,75

a, 357 : x = 5 dư 7

              x = ( 357 - 7 ) : 5

              x = 70

b, x : 4 = 1234 dư 3

    x       = 1234 * 4 + 3

    x        =  4939

                    Hok Tốt

tttttt

1) - 538 - ( - 830 ) = - 1368

2) - 8. (- 30) = - 240

3) -8 . 9 - ( - 48 ) = -72 - ( -48) = - 120

4) - 36 . ( - 264 ) = 9504

5) 111 - 611 = - 500

6) ( - 8 ). ( - 125 ) = 1000

7) 154 + (-54) = 100

8) ( - 134 ) + (-66) = - 200

9) x thuộc { -19, -18,...,20 }

Bạn cố gắng làm hết nhé minh có việc phải đi rồi!

= 3570

= 357x(5-1+6) = 357 x10 = 3570

3570(dễ vãi...)

a: \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)

=>\(\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

b: \(\left|2x+1\right|+\dfrac{3}{2}=2\)

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

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

c: (2x-3)²=36

=>\(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

d: \(7^{x+2}+2\cdot7^x=357\)

=>\(7^x\cdot49+7^x\cdot2=357\)

=>\(7^x=7\)

=>x=1

a) \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

\(---\)

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

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

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

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

\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{3}\\2x=-\dfrac{7}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{7}{6}\end{matrix}\right.\)

\(---\)

c) \(\left(2x-3\right)^2=36\)

\(\Rightarrow\left(2x-3\right)^2=\left(\pm6\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)

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

\(---\)

d) \(7^{x+2}+2\cdot7^x=357\)

\(\Rightarrow7^x\cdot7^2+2\cdot7^x=357\)

\(\Rightarrow7^x\cdot\left(7^2+2\right)=357\)

\(\Rightarrow7^x\cdot\left(49+2\right)=357\)

\(\Rightarrow7^x\cdot51=357\)

\(\Rightarrow7^x=357:51\)

\(\Rightarrow7^x=7\)

\(\Rightarrow x=1\)