(x+1)+(x+2)+(x+3)+...+(x+100)

=(x+x+...)+(1+2+3+4+...+100)

=x.100+(1+100x100:2)

=x.100+5050=5750

x.100=5750-5050

x.100=700

x=700:100

vậy x=7

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

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

          100 số x

<=>   x.100 + ( 100 + 1 ) .100 : 2 = 5750

<=>   x . 100 + 5050 = 5750

=> x . 100 = 700

=> x = 7

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

(x + x + ... + x) + (1 + 2 + 3 + 4 + 5 + ... + 100) = 5750

            x . 100 + (1 + 2 + 3 + 4 + 5 + ... + 100) = 5750

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

                                                x . 100 + 5050 = 5750

                                                           x . 100 = 5750 - 5050

                                                           x . 100 = 700

                                                                    x = 700 : 100 = 7

a) \(S=25x^2-20x+7=\left[\left(5x\right)^2-2.5x.2+4\right]+3=\left(5x-2\right)^2+3>0\) với mọi x

b) \(P=9x^2-6xy+2y^2+1=\left[\left(3x\right)^2-2.3x.y+y^2\right]+y^2+1=\left(3x-y\right)^2+y^2+1>0\)với mọi x

25x²  - 20x + 7 = ( 25x² - 20x + 4 ) + 3 = (5x-2)² + 3 > 0

còn câu b, P = 9x² - 6xy + 2y² + 1 = (3x-y)² + y² + 1 >0

ờ

làm sao hay z bn ✔♬

Ta có: \(x^4-30x^2+31x-30=0\) \(\Rightarrow x^4+x-30x^2+30x-30=0\)

\(\Rightarrow x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)

\(\Rightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)

\(\Rightarrow\left(x^2-x+1\right)\left(x^2+x-30\right)=0\)

Xét \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)

\(\Rightarrow x^2+x-30=0\Rightarrow x^2-5x+6x-30=0\)

\(\Rightarrow\left(x-5\right)\left(x+6\right)=0\Rightarrow\orbr{\begin{cases}x-5=0\\x+6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-6\end{cases}}}\)

Vậy x=5 hoặc x = -6

\(\Leftrightarrow x^4-5x^3+5x^3-25x^3-5x^3+25x+6x-30=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^3+5x^2-5x+6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\cdot\left(x^3+6x^2-x^2-6x+x+6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+6\right)\left(x^2-x+1\right)=0\)

hay \(x\in\left\{5;-6\right\}\)

b) 7/9

\(\text{a)}\dfrac{3}{4}+\dfrac{2}{5}\)

\(=\dfrac{15}{20}+\dfrac{8}{20}=\dfrac{23}{20}\)

\(\text{b)}\dfrac{5}{17}+\dfrac{2}{3}-\dfrac{5}{3}+\dfrac{7}{9}+\dfrac{12}{17}\)

\(=\dfrac{5}{17}+\dfrac{2}{3}+\left(\dfrac{-5}{3}\right)+\dfrac{7}{9}+\dfrac{12}{17}\)

\(=\left(\dfrac{5}{17}+\dfrac{12}{17}\right)+\left[\dfrac{2}{3}+\left(\dfrac{-5}{3}\right)\right]+\dfrac{7}{9}\)

\(=1+\left(-1\right)+\dfrac{7}{9}\)

\(=0+\dfrac{7}{9}=\dfrac{7}{9}\)

\(a,\dfrac{3}{4}+\dfrac{2}{5}\)

\(=\dfrac{15}{20}+\dfrac{8}{20}\)

\(=\dfrac{23}{20}\)

b,

[(x+2)(x+5)][(x+3)(x+4)] -24

= (x²+7x+10)(x²+7x+12) -24

=(x²+7x+11-1)(x²+7x+11+1) -24

=(x²+7x+11)²-1-24

=(x²+7x+11)² -25

=(x²+7x+11-5)(x²+7x+11+5)=(x²+7x+6)(x²+7x+16)

✽

mk nhịu

a)    \(\frac{28\times7-45\times7+7\times18}{45\times14}\)

\(=\frac{7\left(28-45+7\right)}{45\times14}\)

\(=\frac{7\times\left(-10\right)}{45\times14}=\frac{-1}{9}\)

b)   \(\frac{12.3-2.6}{4.5.6}\)

\(=\frac{2.6.3-2.6}{4.5.6}\)

\(=\frac{2.6\left(3-1\right)}{2.2.5.6}\)

\(=\frac{2.6.2}{2.2.5.6}\)\(=\frac{1}{5}\)

a)\(\frac{28\times7-45\times7+7\times18}{45\times14}\)

\(=\frac{7\times\left(28-45+18\right)}{45\times14}\)

\(=\frac{7\times1}{45\times14}\)

\(=\frac{1}{90}\)

b)\(\frac{12\times3-2\times6}{4\times5\times6}\)

\(=\frac{12\times\left(3-1\right)}{4\times5\times6}\)

\(=\frac{6\times2\times2}{2\times2\times5\times6}\)

\(=\frac{1}{5}\)

P/s em tham khảo nhé