\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot....\cdot\frac{30}{62}\cdot\frac{31}{64}=2^x\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot.....\cdot\frac{30}{31}\cdot\frac{31}{32}\right)=2^x\)

\(\Leftrightarrow\frac{1}{32}=2^{x+1}\)

Làm nốt.

ko làm được câu này hay câu b ib với tớ nha.khẳng định tối giải.

Câu b) tạm thời ko bít làm =.= 

Bài 1 : 

\(d)\) \(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2x\)

\(\Leftrightarrow\)\(\frac{4^5.4}{3^5.3}.\frac{6^5.6}{2^5.2}=2x\)

\(\Leftrightarrow\)\(\frac{4^6}{3^6}.\frac{6^6}{2^6}=2x\)

\(\Leftrightarrow\)\(\frac{2^{12}}{3^6}.\frac{2^6.3^6}{2^6}=2x\)

\(\Leftrightarrow\)\(\frac{2^{12}}{3^6}.\frac{3^6}{1}=2x\)

\(\Leftrightarrow\)\(2^{12}=2x\)

\(\Leftrightarrow\)\(x=\frac{2^{12}}{2}\)

\(\Leftrightarrow\)\(x=2^{11}\)

\(\Leftrightarrow\)\(x=2048\)

Vậy \(x=2048\)

Chúc bạn học tốt ~ 

Bài 1 : 

\(a)\) Ta có : 

\(4+\frac{x}{7+y}=\frac{4}{7}\)

\(\Leftrightarrow\)\(\frac{x}{7+y}=\frac{4}{7}-4\)

\(\Leftrightarrow\)\(\frac{x}{7+y}=\frac{-24}{7}\)

\(\Leftrightarrow\)\(\frac{x}{-24}=\frac{7+y}{7}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\frac{x}{-24}=\frac{7+y}{7}=\frac{x+7+y}{-24+7}=\frac{22+7}{-17}=\frac{29}{-17}=\frac{-29}{17}\)

Do đó : 

\(\frac{x}{-24}=\frac{-29}{17}\)\(\Rightarrow\)\(x=\frac{-29}{17}.\left(-24\right)=\frac{696}{17}\)

\(\frac{7+y}{7}=\frac{-29}{17}\)\(\Rightarrow\)\(y=\frac{-29}{17}.7-7=\frac{-322}{17}\)

Vậy \(x=\frac{696}{17}\) và \(y=\frac{-322}{17}\)

Chúc bạn học tốt ~ 

2.

Ta có 1+2+...+n=n.(n+1):2

=>P=\(1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+...+\)\(\frac{1}{16}.\frac{16.17}{2}\)=1+\(\frac{3}{2}+\frac{4}{2}+...+\frac{17}{2}\)=1+\(\frac{1}{2}.\left(3=4+..=17\right)\)

=1+\(\frac{1}{2}.153=1+\frac{153}{2}=\frac{155}{2}\)

b) \(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=\frac{4^5.\left(1+1+1+1\right)}{3^5.\left(1+1+1\right)}.\frac{6^5.\left(1+1+1+1+1+1\right)}{2^5.\left(1+1\right)}\)

                                                                                                       \(=\frac{4^5.4}{3^5.3}.\frac{6^5.6}{2^5.2}=\frac{4^6}{3^6}.\frac{6^6}{2^6}=\frac{2^{12}.2^6.3^6}{3^6.2^6}=2^{12}\)

   Ta có: \(2^{12}=\left(2^3\right)^4=8^4\)                                                    

Vậy x= 4

Bài 1:

\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+1986}\right)\)

Nhận xét: \(1-\frac{1}{1+2+...+n}=1-\frac{2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)

Do đó: \(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+...+1986}\right)\)

\(=\frac{1\cdot4}{2\cdot3}\cdot\frac{2\cdot5}{3\cdot4}\cdot...\cdot\frac{1985\cdot1988}{1986\cdot1987}=\frac{1\cdot4\cdot1988}{1986\cdot3}=\frac{3976}{2979}\)

Bài 2:

\(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}\cdot\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2^x\)

\(\Rightarrow\frac{4\cdot4^5}{3\cdot3^5}\cdot\frac{6\cdot6^5}{2\cdot2^5}=2^x\)\(\Rightarrow\frac{4^6}{3^6}\cdot\frac{6^6}{2^6}=2^x\)

\(\Rightarrow\frac{\left(2^2\right)^6}{3^6}\cdot\frac{\left(2\cdot3\right)^6}{2^6}=2^x\)\(\Rightarrow\frac{2^{12}}{3^6}\cdot\frac{2^6\cdot3^6}{2^6}=2^x\)

\(\Rightarrow\frac{2^6\cdot3^6\cdot2^{12}}{2^6\cdot3^6}=2^x\)\(\Rightarrow2^{12}=2^x\Rightarrow x=12\)

đúng rồi đó bạn ks bạn ý đi chứ

chuẩn luôn

a.4^7

b.8^5 

 c.cho x mk sẻ tính kết quả nhưng tìm xmk ko tính đâu

f: =>35x-5=96-6x

=>41x=101

hay x=101/41

g: =>3(x-3)=90-5(1-2x)

=>3x-9=90-5+10x

=>3x-9=10x+85

=>-7x=94

hay x=-94/7

h.3x - 2/6 - 5 = 3 - 2(x + 7)/4

<=> 3x - 2 - 30/6 = 3 - 2(x + 7)/4

<=> 3x - 32/6 = 3 - 2x - 14/4

<=> 3x - 32/6 = -2x - 11/4

<=> 6x - 64/12 = -6x - 33/12

<=> 6x - 64 = -6x - 33 <=> 12x = 31 <=> x = 31/12

\(=\frac{6\times\frac{5}{3}-\frac{7}{6}\times\frac{6}{7}}{\frac{21}{5}\times\frac{10}{11}+\frac{57}{11}}=\frac{10-1}{\frac{42}{11}+\frac{57}{11}}=\frac{9}{\frac{99}{11}}\)=1

Đây là những bài cơ bản mà bạn!

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

\(< =>\frac{\left(5x-2\right).2}{6}=\frac{\left(5-3x\right).3}{6}\)

\(< =>\left(5x-2\right).2=\left(5-3x\right).3\)

\(< =>10x-4=15-9x\)

\(< =>10x+9x=15+4\)

\(< =>19x=19< =>x=1\)

\(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)

\(< =>\frac{\left(10x+3\right).3}{36}=\frac{36}{36}+\frac{\left(6+8x\right).4}{36}\)

\(< =>\left(10x+3\right).3=36+\left(6+8x\right).4\)

\(< =>30x+9=36+24+32x\)

\(< =>32x-30x=9-36-24\)

\(< =>2x=9-60=-51< =>x=-\frac{51}{2}\)