a, 3600: [ ( 5x + 35 ) : x ] = 50

<=> ( 5x + 35 ) : x = 72

<=> 5x + 35 = 72x

<=> 72x - 5x = 35

<=> 67x = 35

<=> x = 35/67

b, 2^n-1 + 4 . 2ⁿ = 9 . 2⁵ 

<=> 2ⁿ : 2 + 4 . 2ⁿ  = 9 . 2⁵ 

<=> 2ⁿ . 1/2 + 4 . 2ⁿ  = 9 . 2⁵ 

<=> 2ⁿ . ( 1/2 + 4 ) = 9 . 2⁵   

<=> 2ⁿ . 9/2 = 9 . 2⁵ 

<=> 2ⁿ : 2⁵ = 9 : 9/2

<=> 2^{n - 5} = 2¹ 

<=> n - 5 = 1

<=> n = 6

\(B=2019-\frac{2019}{3}-\frac{2019}{6}-\frac{2019}{10}-...-\frac{2019}{45}\)

\(\Leftrightarrow B=2019\left(1-\frac{1}{3}-\frac{1}{6}-\frac{1}{10}-...-\frac{1}{45}\right)\)

\(\Leftrightarrow B=2019\left[1-\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\right)\right]\)

\(\Leftrightarrow B=2019\left[1-\left(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{9.10}\right)\right]\)

\(\Leftrightarrow B=2019\left[1-2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\right]\)

\(\Leftrightarrow B=2019\left[1-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\right]\)

\(\Leftrightarrow B=2019\left[1-2\left(\frac{1}{2}-\frac{1}{10}\right)\right]\)

\(\Leftrightarrow B=2019\left[1-2.\frac{4}{10}\right]\)

\(\Leftrightarrow B=2019\left[1-\frac{4}{5}\right]\)

\(\Leftrightarrow B=2019.\frac{1}{5}\)

\(\Leftrightarrow B=\frac{2019}{5}\)

a) \(\frac{9^{10}.27^7}{81^7.3^{15}}\)= \(\frac{\left(3^2\right)^{10}.\left(3^3\right)^7}{\left(3^4\right)^7.3^{15}}\)= \(\frac{3^{2.10}.3^{3.7}}{3^{4.7}.3^{15}}\)=\(\frac{3^{20}.3^{21}}{3^{28}.3^{15}}\)=\(\frac{3^{41}}{3^{43}}\)= \(\frac{3^{41}}{3^{41}.3^2}\)= \(\frac{1}{3^2}=\frac{1}{9}\)

b) \(\frac{8^3.9^5.27^5}{4^5.81^6}\)= \(\frac{\left(2^3\right)^3.\left(3^2\right)^5.\left(3^3\right)^5}{\left(2^2\right)^5.\left(3^4\right)^6}\)= \(\frac{2^{3.3}.3^{2.5}.3^{3.5}}{2^{2.5}.3^{4.6}}\)=\(\frac{2^9.3^{10}.3^{15}}{2^{10}.3^{24}}\)= \(\frac{2^9.3^{25}}{2^{10}.3^{24}}\)=\(\frac{2^9.3^{24}.3}{2^9.2.3^{24}}\)=\(\frac{3}{2}\)

tra loi nhanh mk dang can

\(A=2008.\left(\frac{1}{2007}-\frac{1000}{1004}\right)-2009.\left(\frac{1}{2007}-2\right)\)

\(\Rightarrow A=2008.\frac{1}{2007}-2000-2009.\frac{1}{2007}+4018\)

\(\Rightarrow A=\frac{1}{2007}.\left(2008-2009\right)-2000+4018\)

\(\Rightarrow A=\frac{1}{2007}.\left(-1\right)-2000+4018\)

\(\Rightarrow A\approx2018\)

Cách 1 : Liệt kê phần tử

VD : Tập hợp A  các STN < 3 : A = { 0 ; 1 ; 2 ; 3}

Cách 2 : Viết tính chất đặc trưng của các phần tử:

VD : A = { x \(\in\)N| x < 3}

Cách 3 : dùng hình vẽ :

* Tia số : 

* Hình : 

a, \(S_1=\left[\left(999-1\right):1+1\right].\frac{\left(1+999\right)}{2}\)\(=499500\)

d, \(S_4=\left[\left(126-24\right):1+1\right].\frac{24+126}{2}\)\(=7725\)

e, \(S_5=\left[\left(79-1\right):3+1\right].\frac{79+1}{2}\)\(=3160\)

\(3x^2+x-6=3\left(x^2+\frac{1}{3}x+\frac{1}{36}\right)-\frac{1}{12}-6=3\left(x+\frac{1}{6}\right)^2-6\frac{1}{12}\)

Ta có \(3\left(x+\frac{1}{6}\right)^2\ge0=>3\left(x+\frac{1}{6}\right)^2-6\frac{1}{12}\ge-6\frac{1}{12}\)

Dấu "=" xảy ra khi \(x+\frac{1}{6}=0=>x=-\frac{1}{6}\)

Vậy ...

\(3x^2+x-6=3\left(x^2+\frac{1}{3}x+\frac{1}{36}\right)-\frac{1}{12}-6=3\left(x+\frac{1}{6}\right)^2-6\frac{1}{12}\)

\(3x^2+x-6=3\left(x^2+\frac{1}{3}x+\frac{1}{36}\right)-\frac{1}{12}-6=3\left(x+\frac{1}{6}\right)^2-6\frac{1}{12}\)

\(3x^2+x-6=3\left(x^2+\frac{1}{3}x+\frac{1}{36}\right)-\frac{1}{12}-6=3\left(x+\frac{1}{6}\right)^2-6\frac{1}{12}\)

ấy mình giải nhầm 

7^5x : 49 = 7²⁸

=> 7^5x : 7² = 7²⁸

=> 7^{5x - 2}    = 28

=>   5x - 2  = 28

=>   5x        = 30

=>     x        = 30 : 5

=>     x        = 6

b) 2^x - 5 = 27

=> 2^x     = 27 + 5

=> 2^x     = 32

=> 2^x     = 2⁵

=>   x     = 5

c) (x⁴)⁵ = x

=> x^4.5 = x

=> x²⁰  = x

=> x²⁰ - x = 0

=> x(x¹⁹ - 1) = 0

=> \(\orbr{\begin{cases}x=0\\x^{19}-1=0\end{cases}}\)=> \(\orbr{\begin{cases}x=0\\x^{19}=1\end{cases}}\)=> \(\orbr{\begin{cases}x=0\\x^{19}=1^{19}\end{cases}}\)=> \(\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

Vậy \(x\in\left\{-1;0;1\right\}\)

\(a,7^{5x}:49=7^{28}\)

\(7^{5x}:7^2=7^{28}\)

\(7^{5x-2}=7^{28}\)

\(\Rightarrow5x-2=28\)\(\Rightarrow5x=30\)\(\Rightarrow x=6\)

\(b,2^x-5=27\)\(\Rightarrow2^x=32\)\(\Rightarrow2^x=2^5\Rightarrow x=5\)

\(c,\left(x^4\right)^5=x\)

\(\Rightarrow x^{20}=x\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

23+347+90+864-1234=90

Câu trả lời:

\(\text{23+347+90+864-1234}=90\)

Chúc bn học tốt!

Ok