a)(4-x)²+15=40

   (4-x)²=40-15

   (4-x)²=25

    (4-x)²=5²

=>4-x=5

x=5-4

x=1

a) ( 4 - x )² + 15 = 40

( 4 - x )² = 40 - 15

( 4 - x )² = 25

( 4 - x )² = 5²

4 - x = 5

     x = 4 - 5

     x = ( - 1 )

b, ( 3x - 2 )¹⁰ = ( 3x - 2 )⁴

=> x \(\in\left\{0;1\right\}\)

c, X = 1 + 4 + 4² + ......... + 4²⁰¹⁷

4X = 4 + 4¹ + ......... + 4²⁰¹⁸

4X - X = ( 4 + 4¹ + ........ + 4²⁰¹⁸ ) - ( 1 + 4 + 4² + ......... + 4²⁰¹⁷ )

4X - X = 4 + 4¹ + ......... + 4²⁰¹⁸ - 1 - 4 - 4² - ......... - 4²⁰¹⁷

=> 3X = 4²⁰¹⁸ - 1

=> X = \(\frac{4^{2018}-1}{3}\)

a) 2x + 1 = 3

2x = 2

x = 1

b) ( 2x - 5 ) + 17 = 6

( 2x - 5 ) = 6 - 17

( 2x - 5 ) = -11

2x = -11 + 5

2x = -6

x = -3

c) 10 - 2 x ( 4 - 3x ) = -4

2 x ( 4 - 3x ) = 14

4 - 3x = 7

3x = -3 

x = -1 

a) 2x + 1 = 3

2x = 3-1

2x=2

x=2:2

x=1

b) ( 2x - 5 ) + 17 = 6

( 2x - 5 ) = 6 - 17

( 2x - 5 ) = -11

2x = -11 + 5

2x = -6

x=-6:2

x = -3

a)\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6.\)

\(\Leftrightarrow x^2-4x+4-x^2+9-6=0\)

\(\Leftrightarrow-4x+7=0\)

\(\Leftrightarrow4x=7\Leftrightarrow x=1,75\)

\(b,4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10.\)

\(\Leftrightarrow4\left(x^2-6x+9\right)-4x^2+1-10=0\)

\(\Leftrightarrow-24x+27=0\)

\(\Leftrightarrow24x=27\Leftrightarrow x=1,125\)

Trả lời :.....................................

\(\Leftrightarrow24x=27\Leftrightarrow x=1,125..........................\)

Hk tốt,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

Học sinh giỏi 6A

$ a/ 12x(x – 5) – 3x(4x - 10) = 120$

`<=>12x^2-60x-12x^2+30x=120`

`<=>-30x=120`

`<=>x=-4`

Vậy `x=-4`

$b/ 9x(x + 4) – 5x(3x + 2) = 112 - 2x(3x + 1)$

`<=>9x^2+36x-15x^2-10x=112-6x^2-2x`

`<=>-6x^2+26x=112-6x^2-2x`

`<=>28x=112`

`<=>x=4`

Vậy `x=4`

$c/ 3x(1 – x) - 5x(3x + 7) = 154 + 9x(5 – 2x)$

`<=>3x-3x^2-15x^2-35x=154+45x-18x^2`

`<=>-32x-18x^2=154+45x-18x^2`

`<=>77x=-154`

`<=>x=-2`

Vậy `x=-2`

1) \(\left|x\right|< 10\)

\(\Leftrightarrow-10< x< 10\)

2) \(\left|x\right|>11\)

\(\Leftrightarrow\left[{}\begin{matrix}x< -11\\x>11\end{matrix}\right.\)

3) \(\left|x\right|\ge2x\left(\forall x\ge0\right)\)

\(\)\(\Leftrightarrow\left[{}\begin{matrix}x\le-2x\\x\ge2x\end{matrix}\right.\) 

\(\Leftrightarrow\left[{}\begin{matrix}3x\le0\\x\le0\end{matrix}\right.\)

\(\Leftrightarrow x=0\) \(\left(thỏa.đk:x\ge0\right)\)

4) \(\left|x\right|\le-3x\left(\forall x\le0\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\left(-3x\right)\\x\le-3x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x\le0\\4x\le0\end{matrix}\right.\)

\(\Leftrightarrow x\le0\) \(\left(thỏa.đk\right)\)

\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{\left(3x-1\right).\left(3x+3\right)}=\frac{3}{10}\)

=> \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{3x-1}-\frac{1}{3x+3}=\frac{3}{10}\)

=> \(\frac{1}{3}-\frac{1}{3x+3}=\frac{3}{10}\)

=> \(\frac{1}{3x+3}=\frac{1}{3}-\frac{3}{10}\)

=> \(\frac{1}{3x+3}=\frac{1}{30}\)

=> 3x + 3 = 30

=> 3.(x + 1) = 30

=> x + 1 = 10

=> x = 9

x = 9 nha bạn