\(\dfrac{x}{2\cdot5}+\dfrac{x}{5\cdot8}+\dfrac{x}{8\cdot11}+\dfrac{x}{11\cdot14}+...+\dfrac{x}{32\cdot35}=\dfrac{33}{70}\)

\(\Leftrightarrow x\left(\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+\dfrac{1}{11\cdot14}+...+\dfrac{1}{32\cdot35}\right)=\dfrac{33}{70}\)

\(\Rightarrow x\cdot\dfrac{1}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+...+\dfrac{3}{32\cdot35}\right)=\dfrac{33}{70}\)

\(\Leftrightarrow x\cdot\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{32}-\dfrac{1}{35}\right)=\dfrac{33}{70}\)

\(\Leftrightarrow x\cdot\dfrac{1}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{35}\right)=\dfrac{33}{70}\)

\(x\cdot\dfrac{1}{3}\cdot\dfrac{33}{70}=\dfrac{33}{70}\)

\(x=\dfrac{33}{70}:\dfrac{33}{70}:\dfrac{1}{3}\)

\(x=3\)

\(M=\dfrac{4n+3}{n-5}=\dfrac{4\left(n-5\right)+23}{n-5}=4+\dfrac{23}{n-5}\)

Để M nguyên =>\(\dfrac{23}{n-5}\) nguyên 

=>\(23⋮n-5\Leftrightarrow n-5\inƯ\left(23\right)\)

\(n-5\in\left\{1;23;-1;-23\right\}\)

\(n\in\left\{6;28;4;-18\right\}\)

a,Có

A = \(x^2-4x+7\)

\(A=\left(x^2-4x+4\right)+3\)

\(A=\left(x-2\right)^2+3\) \(\ge\)3 

Vì \(\left(x-2\right)^2\ge0\forall x\)

Dấu''='' xảy ra khi x=2

Vậy Amin = 3 khi x=2

b, Có

\(B=x^2+10x\)

\(B=\left(x^2+10x+25\right)-25\)

\(B=\left(x+5\right)^2-25\ge-25\) 

Vì \(\left(x+5\right)^2\ge0\forall x\)

Dấu''='' xảy ra khi x=-5

Vậy Bmin = -25 khi x=-5

c,Có

\(C=2x^2-4x+15\)

\(C=2\left(x^2-2x+1\right)+13\)

\(C=2\left(x-1\right)^2+13\ge13\)

Vì \(2\left(x-1\right)^2\ge0\forall x\)

Dấu ''='' xảy ra khi x = 1

Vậy Cmin = 13 khi x=1

d,Có

\(M=x^2-3x+2022\)

\(M=\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{8079}{4}\)

\(M=\left(x-\dfrac{3}{2}\right)^2+\dfrac{8079}{4}\) \(\ge\) \(\dfrac{8079}{4}\)

Vì \(\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\)

Dấu ''='' xảy ra khi x= 3/2

Vậy Mmin = \(\dfrac{8079}{4}\) khi x= 3/2

 

Gọi chiều rộng là x (cm),còn chiều dài là 2.x(cm) 

Theo bài ra ta có:

x*2*x=18

x*x=9

Vì 9 = 3*3 => x=3 ; 2*x =6 

Vậy chiều rộng hình chữ nhật là 3 cm , chiều dài là 6 cm

Chu vi hình chữ nhật đó là: 

(3+6).2 =18 (cm)

 

\(\sqrt{\dfrac{1}{2}x+1}-1=\dfrac{1}{2}\)

\(\sqrt{\dfrac{1}{2}x+1}=\dfrac{1}{2}+1\)

\(\sqrt{\dfrac{1}{2}x+1}=\dfrac{3}{2}\)

\(\sqrt{\dfrac{1}{2}x+1}=\sqrt{\dfrac{9}{4}}\)

=>\(\dfrac{1}{2}x+1=\dfrac{9}{4}\)

\(\dfrac{1}{2}x=\dfrac{5}{4}\)

\(x=\dfrac{5}{2}\)

\(\sqrt{x^2+1}=2\)

\(\Leftrightarrow\sqrt{x^2+1}=\sqrt{4}\)

\(\Rightarrow x^2+1=4\)

\(x^2=3\)

=>\(x=\sqrt{3}\)

Ta có

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{2b}{6}=\dfrac{3c}{12}\)

Áp dụng tính chất dãy tỉ số bằng nhau ,có

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b+3c}{2+6+12}=\dfrac{-20}{20}=-1\)

=>\(\left\{{}\begin{matrix}a=-2\\b=-3\\c=-4\end{matrix}\right.\)

Khi nháp thì mình nghĩ câu b số hạng cuối của tổng  phải là \(2^{x+2015}\)

Vậy sau khi sửa đề mình làm câu b nhé

Ta có \(2^x+2^{x+1}+2^{x+2}+2^{x+3}+...+2^{x+2015}\)

\(=2^x+2^x.2+2^x.2^2+2^x.2^3+...+2^x.2^{2015}\)

\(=2^x\left(1+2+2^2+2^3+...+2^{2015}\right)\)(1)

Đặt \(A=1+2+2^2+2^3+...+2^{2015}\)

=>2A = 2 +2² +2³ +2⁴ +...+2²⁰¹⁶

=>2A-A =A= 2²⁰¹⁶ -1 (2)

Từ (1);(2) =>\(2^x+2^{x+1}+2^{x+2}+2^{x+3}+...+2^{x+2015}\)

= \(2^x.\left(2^{2016}-1\right)\) 

Mà \(2^x+2^{x+1}+2^{x+2}+2^{x+3}+...+2^{x+2005}\) = 2²⁰¹⁹ -8

Nên \(2^x\left(2^{2016}-1\right)=2^{2019}-8\)

=>\(2^x=\dfrac{2^{2019}-8}{2^{2016}-1}=\dfrac{2^3\left(2^{2016}-1\right)}{2^{2016}-1}=2^3\)

=> x=3

Tớ lật lại sách lớp 6 ,thấy bản thân làm như dưới,không biết có được gọi là thuận tiện ko.

\(2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2}}}}=2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{\dfrac{5}{2}}}}=2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{2}{5}}}\) 

\(=2+\dfrac{1}{2+\dfrac{1}{\dfrac{12}{5}}}=2+\dfrac{1}{2+\dfrac{5}{12}}=2+\dfrac{1}{\dfrac{29}{12}}=2+\dfrac{12}{29}=\dfrac{70}{29}\)

 

 

a,\(5^{x-2}-3^2=2^4-\left(2^8.2^4+2^{10}.2^2\right)\)

\(\Leftrightarrow5^{x-2}-3^2=2^4-\left(2^{12}-2^{12}\right)\)

\(\Leftrightarrow5^{x-2}-3^2=2^4\)

\(\Leftrightarrow5^{x-2}=2^4+3^2\)

\(5^{x-2}=25\)

\(5^{x-2}=5^2\)

=> x-2 =2 

=> x=4 b,

b,\(697:\left[\left(15.x+264\right):x\right]=17\)

\(\Rightarrow\left(15.x+364\right):x=697:17\)

\(\left(15.x+364\right):x=41\)

\(\Rightarrow15.x+364=41.x\)

\(\Rightarrow15.x-41.x=-364\)

\(-26x=-364\)

=> x= -364 :-26 = 14

c,\(134:\left(x-3\right)=35+160:5\)

\(134:\left(x-3\right)=35+32\)

\(134:\left(x-3\right)=67\)

=> x-3= 134 :67

x- 3 =2

=> x= 5

d,\(\left(3.x-2^4\right).7^3=2.7^4\)

\(3.x-2^4=2.7^4:7^3\)

\(3.x-2^4=2.\left(7^4:7^3\right)\)

\(3.x-2^4=2.7\)

\(3.x-2^4=14\)

\(3.x=14+2^4\)

3x =30

=> x=10

e,\(\left[\left(10-x\right).2+5\right]:3-2=3\)

\(\left[\left(10-x\right).2+5\right]:3=3+2\)

\(\left[\left(10-x\right).2+5\right]:3=5\)

\(\left(10-x\right).2+5=15\)

\(\left(10-x\right).2=10\)

=> 10-x = 5

=> x =10 -5

x =5