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

2+4+6+...+2x=210

=2(1+2+3+...+x)=210

=1+2+3+...+x=105

=x+1.x/2=105

(x+1)x=210

14.15=210

=>x=14

Vậy x=14

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

 2+4+6+8+10+...+2x=210

=2(1+2+3+4+5+...+x) = 210

=1+2+3+4+5+...+x=105

=x+1*x/2=105

=(x+1)x=210

=14*15=210

=> x=14

ĐÚNG KHÔNG? NẾU ĐÚNG THÌ TICK ??.

đây là toán mà bạn?

Có phải là tiếng anh đâu?

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

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

\(\Leftrightarrow-4x+13=6\)

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

\(\Leftrightarrow x=\frac{7}{4}\)

Vậy \(x=1\).

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

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

\(\Leftrightarrow4x^2-24x+36-4x^2+1=10\)

\(\Leftrightarrow-24x+37=10\)

\(\Leftrightarrow-24x=27\)

\(\Leftrightarrow x=\frac{9}{8}.\)

Mấy pài kia tương tự . :D

cậu khai triển các tích ra là ra thui mà cậu

(x-2)²-(x-3)(x-3)=6

x²-2.x.2+2²-x²-3²=6

(x²-x²)-4x+(2²⁺3²)=6

-4x+13=6

-4x=6-13=-7

x=-7:(-4)=1,75

a, (x-2)^2 - (x-3)(x+3)=6

x^2-4x+4-(x^2-9)=6

x^2-4x+4-x^2+9=6

(x^2-x^2)-4x+13=6

-4x=-7

x=1,75

b, 4(x-3)^2 - (2x-1)(2x+1)=10

4(x^2-6x+9)-(4x^2-1)=10

4x^2-24x+36-4x^2+1=10

-24x+37=10

x=9/8

c,(x-4)^2 - (x+2)(x-2)=6

x^2-8x+16-(x^2-4)=6

x^2-8x+16-x^2+4=6

-8x+20=6

x=7/4

d, 9(x+1)^2 - (3x-2)(3x+2)=10

9(x^2+2x+1)-(9x^2-4)=10

9x^2+18x+9-9x^2+4=10

18x+13=10

x=-1/6

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

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

\(-4x+13=6\)

\(-4x=6-13\)

\(-4x=-7\)

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

\(x=\frac{7}{4}\)

Vậy \(x=\frac{7}{4}\)

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

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

\(4x^2-24x+36-4x^2+1=10\)

\(-24x+37=10\)

\(x=\frac{9}{8}\)

Vậy \(x=\frac{9}{8}\)

\(c,\left(x-4\right)^2-\left(x+2\right)\left(x-2\right)=6\)

\(x^2-8x+16-\left(x^2-4\right)=6\)

\(x^2-8x+16-x^2+4=6\)

\(-8x+20=6\)

\(x=\frac{7}{4}\)

Vậy \(x=\frac{7}{4}\)

\(d,9\left(x+1\right)^2-\left(3x-2\right)\left(3x+2\right)=10\)

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

\(9x^2+18x+9-9x^2+4=10\)

\(18x+13=10\)

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

Vậy \(x=\frac{-1}{6}\)

|2x-4|-|2x-10|=6

_*2x-4\(\ge\)0\(\Leftrightarrow\)2x\(\ge4\)\(\Leftrightarrow x\ge2\):

trường hợp 1:\(2x-10\ge0\Leftrightarrow2x\ge10\Leftrightarrow x\ge5\Rightarrow x\ge5\)ta có:

2x-4-2x-10=6

       2x-2x=6+4+10

            0x=20(vô lý)

trường hợp 2:\(2x-10