\(a,50\%x-0,2+x=\dfrac{4}{5}\)

\(\Leftrightarrow\dfrac{1}{2}x-0,2+x=\dfrac{4}{5}\)

\(\Leftrightarrow\dfrac{1}{2}x+x=\dfrac{4}{5}+0,2\)

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

\(\Leftrightarrow\dfrac{3}{2}x=1\)

\(\Leftrightarrow x=\dfrac{2}{3}\)

\(b,\left(x-\dfrac{3}{4}\right):\dfrac{1}{2}+\dfrac{3}{2}=\dfrac{25}{2}\)

\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{25}{2}-\dfrac{3}{2}\)

\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{22}{2}\)

\(\Leftrightarrow x-\dfrac{3}{4}=11:2\)

\(\Leftrightarrow x=\dfrac{11}{2}+\dfrac{3}{4}\)

\(\Leftrightarrow x=\dfrac{25}{4}\)

Bài lớp \(6\) chưa sử dụng dấu \(\Leftrightarrow\) chị nhé ! Vẫn phải sử dụng dấu \(\Rightarrow\) Khi nào bài lớp \(8\) trở lên thì cj hãy dùng \(\Leftrightarrow\) ạ

\(-2\left(2x-7\right)^2=2\)

\(\Rightarrow\left(2x-7\right)^2=-4\)

Mà: \(\left(2x-7\right)^2\ge0\)

=> Ko có giá trị x cần tìm

Bạn viết lại đề bài đi!  chỗ (2x-+7) có sai không vậy ?

\(A=-5x^3y^2\)

a, 33.( 17- 5) - 17.( 33-5)

= 33.17 - 33.5 - 17.33 + 17.5

= ( 33.17 - 17.33) - ( 33.5 - 17.5)

= 0 - 5.( 33- 17)

= - 5. 16

= - 80

b, 12 + 3.{ 90 : [ 39 - ( 2³ - 5)²]

 = 12 + 3. { 90 : [ 39 - ( 8-5)² ]}

= 12 + 3 . { 90 : [ 39 - 3² ]}

= 12 + 3.{ 90 : (39 -9)}

= 12 + 3. { 90 : 30}

= 12 + 3 . 3

= 12 + 9

= 21

c, 307 - [ (180 .4⁰ - 160 ) : 2² + 9] : 2

= 307 - [ ( 180 - 160) : 4 + 9]:2

= 307 - [ 20:4 +9 ] :2

= 307 - [ 5 + 9] : 2

= 307 - 14 : 2

= 307 - 7

= 300 

a)

`(x+2)^2 -x-2=0`

`<=> x^2 +4x+4-x-2=0`

`<=> x^2+3x+2=0`

`<=> x^2 +2x+x+2=0`

`<=> x(x+2)+(x+2)=0`

`<=> (x+2)(x+1)=0`

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

b)

` c^2 -4c+4=c-2`

`<=> (c-2)^2 -c+2=0`

`<=> (c-2)^2 -(c-2)=0`

`<=> (c-2)(c-2-1)=0`

`<=> (c-2)(c-3)=0`

\(< =>\left[{}\begin{matrix}c-2=0\\c-3=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}c=2\\c=3\end{matrix}\right.\)

Cái Math Processing Error là \(\left[{}\begin{matrix}x=-2\\x=-1\end{matrix}\right.\) bạn nhé.

cảm ơn xong chẳng có ai :)))

Mk xin phép ko vt lại đề nx

\(\Rightarrow A=\left[\left(3x-2\right)\left(x+1\right)-\left(2x+5\right)\left(x^2-1\right)\right]\div x+1\)

\(\Rightarrow A=3x-2-\left(2x-5\right)\left(x-1\right)\)

\(\Rightarrow x=\dfrac{1}{2}\)

\(\Rightarrow A=\dfrac{3}{2}-2-\left(1-5\right)\left(\dfrac{1}{2}-1\right)=-\dfrac{5}{2}\)

Ta có:

\(15x^4y^4-M=10x^2y^4+6x^2y^4\)

\(\Leftrightarrow M=15x^4y^4-\left(10x^2y^4+6x^2y^2\right)\)

\(\Leftrightarrow M=15x^4y^4-16x^2y^4\)

Thay \(x=-\dfrac{1}{2};x=2\) vào M ta có:

\(M=15\cdot\left(-\dfrac{1}{2}\right)^4\cdot2^4-16\cdot\left(-\dfrac{1}{2}\right)^2\cdot2^4=-49\)

A chia hết cho 8

A=(1+7)+7^2(1+7)+......+7^100(1+7)

A=8+7^2.8+.........+7^100.8

A=8(1+7^2+...+7^100) chia hết cho 8

Vậy A chia hết cho 8

A = 1 + 7 + 7² + 7³ +...+ 7¹⁰¹

A = 7⁰ + 7¹ + 7² + 7³ + ... + 7¹⁰¹

A = ( 7⁰ + 7¹ ) + ( 7² + 7³ ) + ... + ( 7¹⁰⁰ + 7¹⁰¹ )

A = 7⁰ . ( 7⁰ + 7¹ ) + 7² . ( 7⁰ + 7¹ ) + ... + 7¹⁰⁰ . ( 7⁰ + 7¹ )

A = 7⁰ . 8 + 7² . 8 + ... + 7¹⁰⁰ . 8

A = 8 . ( 7⁰ + 7² + ... + 7¹⁰⁰ ) \(⋮\)8