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Tính nhanh theo mẫu: Mẫu: \(B=\left(1+\dfrac{1}{3}\right)\)x \(\left(1+\dfrac{1}{8}\right)\)x \(\left(1+\dfrac{1}{15}\right)\)x \(\left(1+\dfrac{1}{24}\right)\)x ..... x \(\left(1+\dfrac{1}{120}\right)\)x \(\left(1+\dfrac{1}{413}\right)\) \(B=\left(\dfrac{3}{3}+\dfrac{1}{3}\right)\)x \(\left(\dfrac{8}{8}+\dfrac{1}{8}\right)\)x \(\left(\dfrac{15}{15}+\dfrac{1}{15}\right)\)x \(\left(\dfrac{24}{24}+\dfrac{1}{24}\right)\)x........x\(\left(\dfrac{120}{120}+\dfrac{1}{120}\right)\)x...
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Tính nhanh theo mẫu:

Mẫu: \(B=\left(1+\dfrac{1}{3}\right)\)x \(\left(1+\dfrac{1}{8}\right)\)x \(\left(1+\dfrac{1}{15}\right)\)x \(\left(1+\dfrac{1}{24}\right)\)x ..... x \(\left(1+\dfrac{1}{120}\right)\)x \(\left(1+\dfrac{1}{413}\right)\)

\(B=\left(\dfrac{3}{3}+\dfrac{1}{3}\right)\)x \(\left(\dfrac{8}{8}+\dfrac{1}{8}\right)\)x \(\left(\dfrac{15}{15}+\dfrac{1}{15}\right)\)x \(\left(\dfrac{24}{24}+\dfrac{1}{24}\right)\)x........x\(\left(\dfrac{120}{120}+\dfrac{1}{120}\right)\)x \(\left(\dfrac{143}{143}+\dfrac{1}{143}\right)\)

\(B=\dfrac{4}{3}\)x\(\dfrac{9}{8}\)x\(\dfrac{16}{15}\)x\(\dfrac{25}{24}\)x.......x\(\dfrac{121}{120}\)x \(\dfrac{144}{143}\)

\(B=\dfrac{2x2}{1x3}\)x\(\dfrac{3x3}{2x4}\)x\(\dfrac{4x4}{3x5}\)x\(\dfrac{5x5}{4x6}\)x.......x\(\dfrac{11x11}{10x12}\)x\(\dfrac{12x12}{13x11}\)

\(B=\dfrac{2x3x4x5x......x10x11x12}{1x2x3x......x10x11x12}\)x \(\dfrac{2x3x4x5x....x11x12}{3x4x5x6x......x12x13}\)

B= \(\dfrac{12}{1}\)x\(\dfrac{2}{13}\)

B=\(\dfrac{24}{13}\)

Câu hỏi:

\(B=\left(1+\dfrac{1}{8}\right)\)x\(\left(1+\dfrac{1}{15}\right)\)x\(\left(1+\dfrac{1}{24}\right)\)x..... x \(\left(1+\dfrac{1}{440}\right)\)x \(\left(1+\dfrac{1}{483}\right)\)

3
24 tháng 6 2017

\(B=\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)\left(1+\dfrac{1}{24}\right).....\left(1+\dfrac{1}{440}\right)\left(1+\dfrac{1}{483}\right)\)

\(B=\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}.....\dfrac{441}{440}.\dfrac{484}{483}\)

\(B=\dfrac{9.16.25.....441.484}{8.15.24.....440.483}\)

\(B=\dfrac{3.3.4.4.5.5.....21.21.22.22}{2.4.3.5.4.6.....20.22.21.23}\)

\(B=\dfrac{3.4.5.....21.22}{2.3.4.....20.21}.\dfrac{3.4.5.....21.22}{4.5.6.....22.23}\)

\(B=11.\dfrac{3}{23}=\dfrac{33}{23}\)

24 tháng 6 2017

B = \(\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}...\dfrac{121}{120}.\dfrac{144}{143}\)

B = \(\dfrac{4.9.16.25...121.144}{3.8.15.24....120.143}\)

B = \(\dfrac{2.2.3.3.4.4.5.5...11.11.12.12}{1.3.2.4.3.5.4.6...10.12.11.13}\)

B = \(\dfrac{2.3.4.5...11.12}{1.2.3.4.5...10.11}.\dfrac{2.3.4.5...11.12}{3.4.5.6.7...12.13}\)

B = 12 . \(\dfrac{2}{13}\)

B = \(\dfrac{24}{13}\)

1: \(MTC=2\left(x-y\right)\left(x+y\right)\)

\(\dfrac{x-y}{2x^2-4xy+2y^2}=\dfrac{x-y}{2\left(x-y\right)^2}=\dfrac{1}{2\left(x-y\right)}=\dfrac{1\cdot\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{x+y}{2\left(x-y\right)\left(x+y\right)}\)

\(\dfrac{x+y}{2x^2+4xy+2y^2}\)

\(=\dfrac{x+y}{2\left(x^2+2xy+y^2\right)}\)

\(=\dfrac{x+y}{2\left(x+y\right)^2}=\dfrac{1}{2\left(x+y\right)}=\dfrac{x-y}{2\left(x+y\right)\left(x-y\right)}\)

\(\dfrac{1}{x^2-y^2}=\dfrac{2}{2\left(x^2-y^2\right)}=\dfrac{2}{2\left(x-y\right)\left(x+y\right)}\)

2: \(\dfrac{1}{x^2+8x+15}=\dfrac{1}{\left(x+3\right)\left(x+5\right)}=\dfrac{x+3}{\left(x+3\right)^2\cdot\left(x+5\right)}\)

\(\dfrac{1}{x^2+6x+9}=\dfrac{1}{\left(x+3\right)^2}=\dfrac{x+5}{\left(x+3\right)^2\cdot\left(x+5\right)}\)

3: \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}=\dfrac{1\cdot\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\dfrac{a-c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(\dfrac{1}{\left(c-b\right)\left(c-a\right)}=\dfrac{1}{\left(b-c\right)\left(a-c\right)}=\dfrac{a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(\dfrac{1}{\left(b-a\right)\left(a-c\right)}=\dfrac{-1}{\left(a-b\right)\left(a-c\right)}=\dfrac{-\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

6:ĐKXĐ: x>=0; x<>1/25

BPT=>\(\dfrac{3\sqrt{x}}{5\sqrt{x}-1}+3< =0\)

=>\(\dfrac{3\sqrt{x}+15\sqrt{x}-5}{5\sqrt{x}-1}< =0\)

=>\(\dfrac{18\sqrt{x}-5}{5\sqrt{x}-1}< =0\)

=>\(\dfrac{1}{5}< \sqrt{x}< =\dfrac{5}{18}\)

=>\(\dfrac{1}{25}< x< =\dfrac{25}{324}\)

7:

ĐKXĐ: x>=0

BPT \(\Leftrightarrow\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}>\dfrac{8}{3}:\dfrac{8}{3}=1\)

=>\(\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}-1>=0\)

=>\(\dfrac{\sqrt{x}+1-2\sqrt{x}-3}{2\sqrt{x}+3}>=0\)

=>\(-\sqrt{x}-2>=0\)(vô lý)

8:

ĐKXĐ: x>=0; x<>9/4

BPT \(\Leftrightarrow\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}+4< 0\)

=>\(\dfrac{\sqrt{x}-2+8\sqrt{x}-12}{2\sqrt{x}-3}< 0\)

=>\(\dfrac{9\sqrt{x}-14}{2\sqrt{x}-3}< 0\)

TH1: 9căn x-14>0 và 2căn x-3<0

=>căn x>14/9 và căn x<3/2

=>14/9<căn x<3/2

=>196/81<x<9/4

TH2: 9căn x-14<0 và 2căn x-3>0

=>căn x>3/2 hoặc căn x<14/9

mà 3/2<14/9

nên trường hợp này Loại

9: 

ĐKXĐ: x>=0

\(BPT\Leftrightarrow\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}< =-\dfrac{1}{3}\)

=>\(\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}+\dfrac{1}{3}< =0\)

=>\(\dfrac{6\sqrt{x}+9+5\sqrt{x}+7}{3\left(5\sqrt{x}+7\right)}< =0\)

=>\(\dfrac{11\sqrt{x}+16}{3\left(5\sqrt{x}+7\right)}< =0\)(vô lý)

10: 

ĐKXĐ: x>=0; x<>1/49

\(BPT\Leftrightarrow\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}+6>0\)

=>\(\dfrac{6\sqrt{x}-2+42\sqrt{x}-6}{7\sqrt{x}-1}>0\)

=>\(\dfrac{48\sqrt{x}-8}{7\sqrt{x}-1}>0\)

=>\(\dfrac{6\sqrt{x}-1}{7\sqrt{x}-1}>0\)

TH1: 6căn x-1>0 và 7căn x-1>0

=>căn x>1/6 và căn x>1/7

=>căn x>1/6

=>x>1/36

TH2: 6căn x-1<0 và 7căn x-1<0

=>căn x<1/6 và căn x<1/7

=>căn x<1/7

=>0<=x<1/49

30 tháng 8 2023

câu 9 nhầm đề bài r bạn

 

1 tháng 5 2018

help me pls!!!

1 tháng 5 2018

giúp bạn cx hơi hảo tổn đó :))

a: =>1+3x-6=-x+3

=>3x-5=-x+3

=>4x=8

=>x=2(loại)

b: \(\Leftrightarrow\dfrac{3\left(x-3\right)+2\left(x-2\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\dfrac{x-1}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)

=>3x-9+2x-4=x-1

=>5x-13=x-1

=>4x=12

=>x=3(loại)

c: =>x^2-2x+4+x^3+8=12

=>x^3+x^2-2x=0

=>x(x^2+x-2)=0

=>x(x+2)(x-1)=0

=>x=0 hoặc x=1

2 tháng 2 2023

tks yeu

20 tháng 1 2022

a, đk x khác 0

<=> x^2 = 16 <=> x = 4 ; x = -4 (tm)

b, <=> 36x +252 = -360 <=> x = -17 

c. đk x khác -1 

<=> (x+1)^2 = 16 

TH1 : x + 1 = 4 <=> x = 3 (tm)

TH2 : x + 1 = -4 <=> x = -5 (tm) 

d, đk x khác 1/2 

<=> (2x-1)^2 = 81 

TH1 : 2x - 1 = 9 <=> x = 5 (tm) 

TH2 : 2x - 1 = -9 <=> x = -4 (tm) 

 

a: \(\Leftrightarrow x^2=16\)

hay \(x\in\left\{4;-4\right\}\)

b: =>x+7/15=-2/3

=>x+7=-10

hay x=-17

c: \(\Leftrightarrow\left(x+1\right)^2=16\)

\(\Leftrightarrow x+1\in\left\{4;-4\right\}\)

hay \(x\in\left\{3;-5\right\}\)

Giải:

a) \(\left(3\dfrac{1}{2}+2x\right).3\dfrac{2}{3}=5\dfrac{1}{3}\) 

     \(\left(\dfrac{7}{2}+2x\right).\dfrac{11}{3}=\dfrac{16}{3}\) 

                 \(\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{11}{3}\) 

                 \(\dfrac{7}{2}+2x=\dfrac{16}{11}\) 

                         \(2x=\dfrac{16}{11}-\dfrac{7}{2}\) 

                         \(2x=\dfrac{-45}{22}\) 

                           \(x=\dfrac{-45}{22}:2\) 

                           \(x=\dfrac{-45}{44}\) 

b) \(3-\left(17-x\right)=-12\) 

       \(3-17+x=-12\) 

                     \(x=-12-3+17\) 

                     \(x=2\) 

c) \(\dfrac{2}{3}x+\dfrac{1}{2}=\dfrac{1}{10}\) 

           \(\dfrac{2}{3}x=\dfrac{1}{10}-\dfrac{1}{2}\) 

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

              \(x=\dfrac{-2}{5}:\dfrac{2}{3}\) 

              \(x=\dfrac{-3}{5}\) 

d) \(\dfrac{3}{4}-2.\left|2x-\dfrac{2}{3}\right|=2\) 

             \(2.\left|2x-\dfrac{2}{3}\right|=\dfrac{3}{4}-2\)  

             \(2.\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{4}\) 

                 \(\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{4}:2\) 

                 \(\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{8}\) 

Vì giá trị tuyệt đối của 1 số nguyên ko bao giờ là số âm nên \(x\in\varnothing\) 

e) \(\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}-\left(-1\right)=\dfrac{1}{3}\) 

                \(\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}=\dfrac{1}{3}+\left(-1\right)\) 

                \(\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}=\dfrac{-2}{3}\) 

                           \(-0,6x-\dfrac{1}{2}=\dfrac{-2}{3}:\dfrac{3}{4}\) 

                           \(-0,6x-\dfrac{1}{2}=\dfrac{-8}{9}\) 

                                   \(-0,6x=\dfrac{-8}{9}+\dfrac{1}{2}\) 

                                   \(-0,6x=\dfrac{-7}{18}\) 

                                           \(x=\dfrac{-7}{18}:-0.6\) 

                                           \(x=\dfrac{35}{54}\) 

f) \(\left(3x-1\right).\left(\dfrac{-1}{2}x+5\right)=0\) 

\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\\dfrac{-1}{2}x+5=0\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\) 

g) \(60\%.x+\dfrac{2}{3}=\dfrac{1}{3}.6\dfrac{1}{3}\) 

        \(\dfrac{3}{5}.x+\dfrac{2}{3}=\dfrac{1}{3}.\dfrac{19}{3}\) 

        \(\dfrac{3}{5}.x+\dfrac{2}{3}=\dfrac{19}{9}\) 

               \(\dfrac{3}{5}.x=\dfrac{19}{9}-\dfrac{2}{3}\) 

               \(\dfrac{3}{5}.x=\dfrac{13}{9}\) 

                    \(x=\dfrac{13}{9}:\dfrac{3}{5}\) 

                   \(x=\dfrac{65}{27}\) 

Chúc bạn học tốt!

f)câu khó nhất

=>3x-1=0 và -1/2x+5=0

   =>x=1/3 và x=10

21 tháng 4 2017

Giải bài 32 trang 50 Toán 8 Tập 1 | Giải bài tập Toán 8