A.   \(\left(x+1\right)+\left(x+2\right)+......+\left(x+100\right)=5750\)

      \(x+1+x+2+....+x+100=5750\)

      \(100x+\left(1+2+3+.......+100\right)=5750\)

      \(100x+5050=5750\)

\(100x=700\)

\(x=700:100=7\)

B.   x+(1+2+......+100) = 2000

       x + 5050 = 2000

            x = 2000 - 5050

           x= -3050

C.   ( x-1 )+(x-2)+......+( x - 100 ) = 50

 x-1+x-2+.........+x-100 = 50

100x + ( -1-2-........-100  ) = 50

100x + ( - 5050 ) = 50

100x = 50 + 5050

100 x = 5100

x = 5100 : 100

x = 51

A . \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)

\(\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)

\(100x+5050=5750\)

\(100x=5750-5050\)

\(100x=700\)

\(\Rightarrow x=\frac{700}{100}=7\)

B. \(x+\left(1+2+3+4+5+....+100\right)=2000\)

 \(x+\frac{\left(100+1\right).100}{2}=2000\)

\(x+5050=2000\)

\(\Rightarrow x=2000-5050=-3050\)

C. \(\left(x-1\right)+\left(x-2\right)+\left(x-3\right)+....+\left(x-100\right)=50\)

\(\left(x+x+x+...+x\right)-\left(1+2+3+...+100\right)=50\)

\(100x-5050=50\)

\(100x=5100\)

\(\Rightarrow x=\frac{5100}{100}=51\)

đặt   \(x^2+x+2\)  là a  ; đặt  \(x+1\)là b

\(\Rightarrow a+b=x^2+x+2+x+1\)\(=x^2+2x+3\)

\(\Rightarrow a^3+b^3=\left(a+b\right)^3\)

\(\Rightarrow a^3+b^3=a^3+3a^2b+3ab^2+b^3\)

\(\Rightarrow3a^2b+3ab^2=0\)\(\Rightarrow3ab\left(a+b\right)=0\)\(\Rightarrow\)\(a=0\)hoặc \(b=0\)hoặc \(a+b=0\)

* nếu a = 0  \(\Rightarrow\) \(x^2+x+2=0\)( vô lí vì luôn dương, cái này dễ chứng minh nha)

* nếu b = 0   \(\Rightarrow x+1=0\Rightarrow x=-1\)

* nếu a + b = 0 \(\Rightarrow x^2+2x+3=0\)(cái này cũng luôn dương nhé)

Vậy phương trình có 1 nghiệm là x = -1 

chúc bạn học tốt nha <3

Thanks bạn nhìu

Ta có : \(\left(3x-2\right)\left(4x+3\right)=\left(2-3x\right)\left(x-1\right)\)

\(\Leftrightarrow12x^2-8x+9x-6=2x-3x^2-2+3x\)

\(\Leftrightarrow12x^2-8x+9x-6-2x+3x^2+2-3x=0\)

\(\Leftrightarrow15x^2-4x-4=0\)​

\(\Leftrightarrow15x^2-10x+6x-4=0\)

Lỗi :vvvv

\(\Leftrightarrow10x\left(\dfrac{3}{2}x-1\right)+4\left(\dfrac{3}{2}x-1\right)=0\)

\(\Leftrightarrow\left(10x+4\right)\left(\dfrac{3}{2}x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy ...

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

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

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

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

=(3x-2)(5x+2)

=15x²-10x-4+6x=15x²-4x-4

\(x\) x 2 + \(\dfrac{x}{2}\)= 10

=> \(x\) x 2 + \(x\) x \(\dfrac{1}{2}\)= 10

=> \(x\)x ( 2 + \(\dfrac{1}{2}\)) = 10

=> \(x\) x \(\dfrac{5}{2}\) = 10

=> \(x\)= 10 : \(\dfrac{5}{2}\)

=> \(x\) = 4

=>5/2x=10

hay x=4

X=1 

\(\frac{x-3}{x-2}+\frac{x-2}{x-4}=3\frac{1}{5}\)

\(=\frac{x-3}{x-2}+\frac{x-2}{x-4}=\frac{16}{5}\)

\(\Rightarrow5\left(x-3\right)\left(x-4\right)+5\left(x-2\right)\left(x-2\right)=16\left(x-2\right)\left(x-4\right)\)

\(\Leftrightarrow5x^2-35x+60+5x^2-20x+20=16x^2-96x+128\)

\(\Leftrightarrow10x^2-55x+80=16x^2-96x+128\)

\(\Leftrightarrow-6x^2+41x-48=0\)

......

\(\frac{x-3}{x-2}+\frac{x-2}{x-4}=3\frac{1}{5}\)

\(\Leftrightarrow\frac{x-3}{x-2}+\frac{x-2}{x-4}=\frac{16}{5}\)

\(\Leftrightarrow\frac{5\left(x-3\right)\left(x-4\right)+5\left(x-2\right)^2}{5\left(x-2\right)\left(x-4\right)}=\frac{16.\left(x-2\right)\left(x-4\right)}{5\left(x-2\right)\left(x-4\right)}\)

\(\Rightarrow5x^2-20x-15x+60+5x^2-20x+20=16x^2-64x-32x+128\)

\(\Leftrightarrow10x^2-55x+80=16x^2-96x+128\)

\(\Leftrightarrow6x^2-41x+48=0\)

\(\Leftrightarrow x=\frac{16}{3};x=\frac{3}{2}\)

\(x^2-4x-1=0\)

\(\left(x^2-2\cdot x\cdot2+4\right)-5=0\)

\(\left(x-2\right)^2=\left(\sqrt{5}\right)^2\)

\(\Rightarrow x-2=\pm\sqrt{5}\)

Tự giải tiếp nha ...

\(x^2-4x-1=0\)

\(\Delta=\left(-4\right)^2-4.\left(-1\right)=20\)

pt có 2 nghiệm

\(x_1=\frac{-4-\sqrt{20}}{2}=-2-\sqrt{5}\)

\(x_2=\frac{-4+\sqrt{20}}{2}=-2+\sqrt{5}\)