\(x\cdot\left(x+7\right)=0\)

Trường hợp 1:

\(x=0\)

Trường hợp 2:

\(x+7=0\)

\(\Rightarrow x=-7\)

Vậy: \(x\in\left\{0;-7\right\}\)

\(x\times\left(x+7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+7=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=0-7=-7\end{matrix}\right.\)

Vậy...

\(x:\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)=100\)

\(\Rightarrow x:\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...+\dfrac{1}{99}-\dfrac{1}{100}\right)=100\)

\(\Rightarrow x:\left(1-\dfrac{1}{100}\right)=100\)

\(\Rightarrow x:\dfrac{99}{100}=100\)

\(\Rightarrow x=100\cdot\dfrac{99}{100}\)

\(\Rightarrow x=99\)

\(x:\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)=100\)

\(x:\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=100\)

\(x:\left(\dfrac{1}{1}-\dfrac{1}{100}\right)=100\)

\(x:\left(\dfrac{100}{100}-\dfrac{1}{100}\right)=100\)

\(x:\dfrac{99}{100}=100\)

\(x=100\cdot\dfrac{99}{100}\)

\(x=99\)

\(-45:5\cdot\left(-3-2x\right)=3\)

\(\Rightarrow-9\cdot\left(-3-2x\right)=3\)

\(\Rightarrow-3-2x=-9:3\)

\(\Rightarrow-3-2x=-3\)

\(\Rightarrow-2x=-3+3\)

\(\Rightarrow-2x=0\)

\(\Rightarrow x=0:\left(-2\right)\)

\(\Rightarrow x=0\)

Ta có công thức tổng quát của dãy số: 

\(1\cdot2+2\cdot3+3\cdot4+...+n\left(n+1\right)=\dfrac{n\left(n+1\right)\left(n+2\right)}{3}\)

Trong đề bài ta có dãy số: \(1\cdot2+2\cdot3+...+99\cdot100\) có \(n=99\)

\(\Rightarrow1\cdot2+2\cdot3+...+99\cdot100=\dfrac{99\cdot\left(99+1\right)\cdot\left(99+2\right)}{3}=333300\)

Trở lại để bài:

\(\dfrac{1\cdot2+2\cdot3+3\cdot4+...+99\cdot100}{x^2+\left(x^2+1\right)+\left(x^2+2\right)+...+\left(x^2+99\right)}=50\dfrac{116}{131}\)

\(\Rightarrow\dfrac{333300}{x^2+x^2+1+x^2+2+...+x^2+99}=\dfrac{6666}{131}\)

\(\Rightarrow\dfrac{333300}{\left(x^2+x^2+...+x^2\right)+\left(1+2+3+...+99\right)}=\dfrac{6666}{131}\)

\(\Rightarrow\dfrac{333300}{100x^2+4950}=\dfrac{6666}{131}\)

\(\Rightarrow6666\left(100x^2+4950\right)=333300\cdot131\)

\(\Rightarrow666600x^2+32996700=43662300\)

\(\Rightarrow666600x^2=10665600\)

\(\Rightarrow x^2=\dfrac{10665600}{666600}\)

\(\Rightarrow x^2=16\)

\(\Rightarrow x^2=4^2\)

\(\Rightarrow x=\pm4\)

Ta có: 1+2-3-4+5+6-7-8+9+...+994-995-996+997+998

= (1-3+5-7+...+993-995+997) + (2-4+6-8+...994-996+998)

= (-2-2-2-2...-2+997) + (-2-2-2...-2)

= 499 + 500

= 999

Ta sử dụng phương pháp đánh giá

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

\(\Rightarrow\left(x-1\right)^2=6-5y^2\)

Ta có: \(\left(x-1\right)^2\ge0\forall x\) 

\(\Rightarrow6-5y^2\ge0\forall y\) (vế trái luôn lớn hơn hoặc bằng 0 thì vế phải cũng vậy) 

\(\Rightarrow5y^2\le6\)

\(\Rightarrow y^2\le1,2\)

Do \(y^2\) là một số nguyên bình phương nên \(\Rightarrow y^2\in\left\{1;0\right\}\Rightarrow y\in\left\{0;1;-1\right\}\)

Thay \(y=0\) vào ta có: \(\left(x-1\right)^2+5\cdot0^2=6\Rightarrow\left(x-1\right)^2=6\) (x không có giá trị nguyên) 

Thay \(y=1\) vào ta có: \(\left(x-1\right)^2+5\cdot1^2=6\Rightarrow\left(x-1\right)^2=1\)

TH1: \(x-1=1\Rightarrow x=2\)

TH2: \(x-1=-1\Rightarrow x=0\)

Thay \(y=-1\) vào ta có: \(\left(x-1\right)^2+5\cdot\left(-1\right)^2=6\Rightarrow\left(x-1\right)^2=1\)

TH1: \(x=2\)

TH2: \(x=0\) 

Vậy: \(\left(x;y\right)=\left\{\left(2;1\right);\left(0;1\right);\left(2;-1\right);\left(0;-1\right)\right\}\)

(\(x\) - 1)² + 5y² = 6 Vì 5y²≥ 0 ⇒ (\(x-1\))² ≤ 6 - 0 = 6

⇒ \(\left[{}\begin{matrix}\left(x-1\right)^2=0;y^2=\dfrac{6}{5}\left(ktm\right)\\\left(x-1\right)^2=1;y^2=\dfrac{6-1}{5}=1\\\left(x-1\right)^2=4;y^2=\dfrac{6-4}{5}=\dfrac{2}{5}\left(ktm\right)\end{matrix}\right.\)

Lập bảng ta có:

Theo bảng trên ta có các cặp \(x;y\) nguyên thỏa mãn đề bài là:

(\(x;y\)) = (0; -1); (0; 1); (2; -1); (2; 1)

\(1\dfrac{13}{15}\times\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{1}{4}\right):\dfrac{7}{5}\)

\(=\dfrac{28}{15}\times\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{5}{20}\right):\dfrac{7}{5}\)

\(=\dfrac{7}{5}\times1-\dfrac{16}{20}:\dfrac{7}{5}\)

\(=\dfrac{7}{5}-\dfrac{16}{20}\times\dfrac{5}{7}\) 

\(=\left(\dfrac{7}{5}\times\dfrac{5}{7}\right)-\dfrac{16}{20}\)

\(=\dfrac{1}{1}-\dfrac{16}{20}\)

\(=1-\dfrac{4}{5}\)

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

Sửa lại, xin lỗi bạn, bạn thông cảm giúp mình nhé!

\(1\dfrac{13}{15}\times\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{1}{4}\right):\dfrac{7}{5}\)

\(=\dfrac{28}{15}\times\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{5}{20}\right)\times\dfrac{5}{7}\)

\(=\dfrac{7}{5}\times1-\dfrac{16}{20}\times\dfrac{5}{7}\)

\(=\dfrac{7}{5}-\dfrac{80}{140}\)

\(=\dfrac{7}{5}-\dfrac{4}{7}\)

\(=\dfrac{49}{35}-\dfrac{20}{35}\)

\(=\dfrac{29}{35}\)

a) \(2x-\dfrac{2}{11}=1\dfrac{1}{5}\)

\(\Rightarrow2x=\dfrac{6}{5}+\dfrac{2}{11}\)

\(\Rightarrow2x=\dfrac{76}{55}\)

\(\Rightarrow x=\dfrac{76}{55}:2\)

\(\Rightarrow x=\dfrac{38}{55}\) 

b) \(\dfrac{5}{9}+\dfrac{4}{3}x=\dfrac{-1}{3}\)

\(\Rightarrow\dfrac{4}{3}x=\dfrac{-1}{3}-\dfrac{5}{9}\)

\(\Rightarrow\dfrac{4}{3}x=-\dfrac{8}{9}\)

\(\Rightarrow x=-\dfrac{8}{9}:\dfrac{4}{3}\)

\(\Rightarrow x=-\dfrac{2}{3}\)

c) \(\dfrac{-3}{7}-\dfrac{3}{5}x=\dfrac{4}{5}+\dfrac{-2}{3}\)

\(\Rightarrow\dfrac{-3}{7}-\dfrac{3}{5}x=\dfrac{2}{15}\)

\(\Rightarrow\dfrac{3}{5}x=-\dfrac{3}{7}-\dfrac{2}{15}\)

\(\Rightarrow\dfrac{3}{5}x=\dfrac{-59}{105}\)

\(\Rightarrow x=\dfrac{-59}{105}:\dfrac{3}{5}\)

\(\Rightarrow x=-\dfrac{59}{63}\)

d) \(\dfrac{3x}{8}=\dfrac{-6}{15}\cdot\dfrac{5}{14}\)

\(\Rightarrow\dfrac{3x}{8}=\dfrac{-3}{7}\)

\(\Rightarrow8\cdot\dfrac{3x}{8}=8\cdot\dfrac{-3}{7}\)

\(\Rightarrow3x=-\dfrac{24}{7}\)

\(\Rightarrow x=\dfrac{-24}{7}:3\)

\(\Rightarrow x=-\dfrac{8}{7}\)

Lời giải:
Gọi thương trong hai phép chia bằng nhau và bằng $a$. 

Số cần tìm là $30a+16=32a+8$

$\Rightarrow 2a=8$

$\Rightarrow a=4$

Số cần tìm là: $30\times 4+16=136$

Ta có:

\(30a+16=32a+8\)

\(2a=8->a=4\)

Số cần tìm là:

\(30\times4+16=136\)

Đáp số: \(136\)

9.

\(A-1=\frac{17}{10^{12}-11}\\ B-1=\frac{17}{10^{11}-12}\)

Mà $10^{12}-11> 10^{12}-12> 10^{11}-12$

$\Rightarrow \frac{17}{10^{12}-11}< \frac{17}{10^{11}-12}$

$\Rightarrow A-1< B-1$

$\Rightarrow A< B$

8.

\(10A=\frac{10^{2013}+10}{10^{2013}+1}=1+\frac{9}{10^{2013}+1}> 1+\frac{9}{10^{2014}+1}=\frac{10^{2014}+10}{10^{2014}+1}=10B\)

$\Rightarrow A>B$