a)  (x+1) +(x+2) +...+(x+100) = 5750

100.x + (1+2+...+100) =5750

100.x + (100.101):2 = 5750

100.x + 5050 = 5750

100x= 700

x= 7

`x-3/5 =-4/10`

`=>x=-4/10 + 3/5`

`=>x=-4/10 + 6/10`

`=>x= 2/10`

`=>x=1/5`

`----`

`(x-1)/(-2) = -8/(x-1)`

`=> (x-1).(x-1) = (-2).(-8)`

`=> (x-1)^2 = 16`

`=> (x-1)^2 = +- 4^2`

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

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

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

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

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

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

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

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

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

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

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

\(A=x^2-2x+10\)

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

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

Mà  \(\left(x-1\right)^2\ge0\)

\(\Rightarrow A\ge9\)

Dấu "=" xảy ra khi :

\(x-1=0\Leftrightarrow x=1\)

Vậy Min A = 9 khi x = 1

\(B=x^2-5x-7\)

\(B=\left(x^2-5x+\frac{25}{4}\right)-\frac{53}{4}\)

\(B=\left(x-\frac{5}{2}\right)^2-\frac{53}{4}\)

Mà  \(\left(x-\frac{5}{2}\right)^2\ge0\)

\(\Rightarrow B\ge-\frac{53}{4}\)

Dấu "=" xảy ra khi :

\(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)

Vậy  \(B_{Min}=-\frac{53}{4}\Leftrightarrow x=\frac{5}{2}\)

Bài 1:

a) \(x.\dfrac{3}{4}=\dfrac{9}{14}\)

\(\Rightarrow x=\dfrac{9}{14}:\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{6}{7}\)

b) \(x:\dfrac{5}{9}=\dfrac{3}{10}\)

\(\Rightarrow x=\dfrac{3}{10}.\dfrac{5}{9}\)

\(\Rightarrow x=\dfrac{1}{6}\)

help me

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

cho hỏi 11 x với gì vậy

a )Ta có :

|x + 1| ≥ 0

|x + 2| ≥ 0

|x + 3| ≥ 0

.......

|x + 10| ≥ 0

=> |x + 1| + |x + 2 | + .... + |x + 10| ≥ 0

Mà |x + 1| + |x + 2 | + .... + |x + 10| = 11x

=> 11x ≥ 0 => x ≥ 0

Vì x ≥ 0 => |x + 1| + |x + 2 | + .... + |x + 10| = x + 1 + x + 2 + ... + x + 10 = 11x

=> 10x + 55 = 11x => 11x - 10x = 55 => x = 55

b ) 2(x + 1) - 4(x - 3) = 2017

<=> 2x + 2 - 4x + 12 = 2017

<=> 2x - 4x + 2 + 12 = 2017

<=> - 2x + 14 = 2017

<=> -2x = 2003

=> x = -1001.5

\(6,\\ a,\\ 1,A=x^2+3x+7=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)

Dấu \("="\Leftrightarrow x=-\dfrac{3}{2}\)

\(2,B=\left(x-2\right)\left(x-5\right)\left(x^2-7x+10\right)=\left(x-2\right)^2\left(x-5\right)^2\ge0\)

Dấu \("="\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

\(b,\\ 1,A=11-10x-x^2=-\left(x+5\right)^2+36\le36\)

Dấu \("="\Leftrightarrow x=-5\)

cảm ơn nha:3

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 \(x=\frac{-1}{6}\)