\(a,\Rightarrow2x^2-18x-2x^2=0\\ \Rightarrow-18x=0\Rightarrow x=0\\ b,\Rightarrow2x^2-5x-12+x^2-7x+10=3x^2-17x+20\\ \Rightarrow5x=22\Rightarrow x=\dfrac{22}{5}\)

Ta có: \(\frac{3}{1\cdot2}+\frac{3}{3\cdot4}+\cdots+\frac{3}{299\cdot300}\)

\(=3\cdot\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{299\cdot300}\right)\)

\(=3\left(1-\frac12+\frac13-\frac14+\cdots+\frac{1}{299}-\frac{1}{300}\right)\)

\(=3\cdot\left\lbrack1+\frac12+\frac13+\frac14+\cdots+\frac{1}{299}+\frac{1}{300}-2\left(\frac12+\frac14+\cdots+\frac{1}{300}\right)\right\rbrack\)

\(=3\left(1+\frac12+\frac13+\cdots+\frac{1}{300}-1-\frac12-\cdots-\frac{1}{150}\right)\)

\(=3\left(\frac{1}{151}+\frac{1}{152}+\cdots+\frac{1}{300}\right)\)

Ta có: \(\left(\frac{3}{1\cdot2}+\frac{3}{3\cdot4}+\cdots+\frac{3}{299\cdot300}\right)\cdot\left(x+\frac23\right)=\frac{2}{151}+\frac{2}{152}+\cdots+\frac{2}{300}\)

=>\(3\left(\frac{1}{151}+\frac{1}{152}+\cdots+\frac{1}{300}\right)\left(x+\frac23\right)=2\left(\frac{1}{151}+\frac{1}{152}+\cdots+\frac{1}{300}\right)\)

=>\(x+\frac23=\frac23\)

=>x=0

=>3(x-2)=4(x+5)

=>4x+20=3x-6

=>x=-26

`(x-2):4 = (5+x) :3`

`(x-2)*3 = 4*(5+x)`

`3x -6 = 20 +4x`

`3x -4x = 20 +6`

`-x =26`

`x=26`

Thời gian có hạn copy cái này hộ mình vào google xem nha: :

Link :   https://lazi.vn/quiz/d/16491/nhac-edm-la-loai-nhac-the-loai-gi

Vào xem xong các bạn nhận được 1 thẻ cào mệnh giá 100k nhận thưởng bằng cách nhắn tin vs mình và 1 phần thưởng bí mật là chiếc áo đá bóng,....

Có 500 giải nhanh nha đã có 200 người nhận rồi

OK

Vao đi

(x+9)+(x-2)+(x+7)+(x-4)+(x+5)+(x-6)+(x+3)+(x-8)+(x+1)=95

x + 9 + x – 2 + x + 7 + x – 4 + x + 5 + x – 6 + x + 3 + x – 8 + x + 1 = 95

x × 9 + (9 - 8) + (7 - 6) + (5 - 4) + (3 - 2) + 1= 95

x × 9 + 5 = 95

x × 9 = 90

x = 10 

a, \(x=\left(\dfrac{3}{5}+\dfrac{2}{5}\right):\dfrac{4}{5}=\dfrac{5}{4}\)

b, \(x=\left(-\dfrac{1}{2}-\dfrac{2}{5}\right).\dfrac{3}{4}=-\dfrac{27}{40}\)

a)\(x=\left(\dfrac{3}{5}+\dfrac{2}{5}\right):\dfrac{4}{5}=\dfrac{5}{4}\)

b)\(x=\left(-\dfrac{1}{2}-\dfrac{2}{5}\right)\times\dfrac{3}{4}=-\dfrac{27}{40}\)

\(1,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 2,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 3,\left(x+1\right)^2+2\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x+1+2\right)=0\\ \Rightarrow\left(x+1\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

x²+4x+4=0
(x+2)²=0
x+2=0
x=+-2
câu 1 giống câu 2
(x+1)²+2(x+1)=0
(x+1+2)(x+1)=0
Th1: x+3=0           Th2: x+1=0
            x=-3                      x=-1
vậy ...

\(=\dfrac{11}{4}:\dfrac{33}{16}-0,5+\left(\dfrac{14}{5}-3\right)^2\\ =\dfrac{11}{4}\cdot\dfrac{16}{33}-\dfrac{1}{2}+\left(-\dfrac{1}{5}\right)^2\\ =\dfrac{4}{3}-\dfrac{1}{2}+\dfrac{1}{25}=\dfrac{131}{150}\)