1+1/3+1/9+1/27+1/81+1/243+1/720=?
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Để chứng minh rằng √(a-b) và √(3a+3b+1) là các số chính phương, ta sẽ điều chỉnh phương trình ban đầu để tìm mối liên hệ giữa các biểu thức này. Phương trình ban đầu: 2^(2+a) = 3^(2+b) Ta có thể viết lại phương trình theo dạng: (2^2)^((1/2)+a/2) = (3^2)^((1/2)+b/2) Simplifying the exponents, we get: 4^(1/2)*4^(a/2) = 9^(1/2)*9^(b/2) Taking square roots of both sides, we have: √4*√(4^a) = √9*√(9^b) Simplifying further, we obtain: 22*(√(4^a)) = 32*(√(9^b)) Since (√x)^y is equal to x^(y/), we can rewrite the equation as follows: 22*(4^a)/ = 32*(9^b)/ Now let's examine the expressions inside the square roots: √(a-b) can be written as (√((22*(4^a))/ - (32*(9^b))/)) Similarly, √(3*a + 3*b + ) can be written as (√((22*(4^a))/ + (32*(9^b))/)) We can see that both expressions are in the form of a difference and sum of two squares. Therefore, it follows that both √(a-b) and √(3*a + 3*b + ) are perfect squares.
\(p=\left[\left(x+5\right).\left(x+11\right)\right].\left[\left(x+7\right).\left(x+9\right)\right]+16=\)
\(=\left(x^2+16x+55\right)\left(x^2+16x+63\right)+16=\)
\(=\left(x^2+16x\right)^2+118.\left(x^2+16x\right)+3481=\)
\(=\left(x^2+16x\right)^2+2.\left(x^2+16x\right).59+59^2=\)
\(=\left[\left(x^2+16x\right)+59\right]^2\) là một số chính phương
\(a,3\left|2x-4\right|-5=7\\ 3\left|2x-4\right|=12\\ \left|2x-4\right|=4\\ \left|2x-4\right|=\left(\pm2\right)^2\\ \Rightarrow\left[{}\begin{matrix}2x-4=2\\2x-4=-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=6\\2x=2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy x={3;1}
\(a,3\left|2x-4\right|-5=7\\ \Rightarrow3\left|2x-4\right|=12\\ \Rightarrow\left|2x-4\right|=4\\ \Rightarrow\left[{}\begin{matrix}2x-4=4\\2x-4=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=8\\2x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)
Vậy x={4;0}
\(Q=\left(a^2b^2+a^2+b^2+1\right)\left(c^2+1\right)=\)
\(=a^2b^2c^2+a^2b^2+a^2c^2+a^2+b^2c^2+b^2+c^2+1=\)
\(=a^2b^2c^2+\left(a^2b^2+b^2c^2+a^2c^2\right)+\left(a^2+b^2+c^2\right)+1\) (1)
Ta có
\(\left(ab+bc+ac\right)^2=a^2b^2+b^2c^2+a^2c^2+2ab^2c+2abc^2+2a^2bc=\)
\(=a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=1\)
\(\Rightarrow a^2b^2+b^2c^2+a^2c^2=1-2abc\left(a+b+c\right)\) (2)
Ta có
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ac\right)=\)
\(=a^2+b^2+c^2+2\)
\(\Rightarrow a^2+b^2+c^2=\left(a+b+c\right)^2-2\) (3)
Thay (2) và (3) vào (1)
\(Q=a^2b^2c^2+1-2abc\left(a+b+c\right)+\left(a+b+c\right)^2-2+1=\)
\(=\left(abc\right)^2-2abc\left(a+b+c\right)+\left(a+b+c\right)^2=\)
\(=\left[abc-\left(a+b+c\right)\right]^2\)
Số bi chia 89; thương là 9
Nên \(89:9=9\) dư \(89-9x9=89-81=8\)
Vậy Số chia là 9 , số dư là 8 thỏa đề bài
So sánh
\(A=\dfrac{1999^{1999}+1}{1999^{1998}+1}\) ; \(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}\)
Ta có: \(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}>1\) ( vì tử > mẫu )
Do đó: \(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}>\dfrac{1999^{2000}+1+1998}{1999^{1999}+1+1998}=\dfrac{1999^{2000}+1999}{1999^{1999}+1999}=\dfrac{1999.\left(1999^{1999}+1\right)}{1999.\left(1999^{1998}+1\right)}=\dfrac{1999^{1999}+1}{1999^{1998}+1}=A\)
Vậy B > A
Chúc bạn học tốt
\(\dfrac{5}{8}x-\dfrac{1}{3}x-\dfrac{1}{6}x=15\)
\(\Rightarrow x\left(\dfrac{5}{8}-\dfrac{1}{3}-\dfrac{1}{6}\right)=15\)
\(\Rightarrow x\left(\dfrac{15}{24}-\dfrac{8}{24}-\dfrac{4}{24}\right)=15\)
\(\Rightarrow x.\dfrac{3}{24}=15\)
\(\Rightarrow x.\dfrac{1}{8}=15\Rightarrow x=15:\dfrac{1}{8}=15.\dfrac{8}{1}=120\)
\(81\cdot\left(-\dfrac{4}{3}\right)^2-25\cdot\left(-\dfrac{3}{5}\right)^3-3^3\\ =81\cdot\dfrac{16}{9}-25\cdot-\dfrac{27}{125}-27\\ =144--\dfrac{27}{5}-27\\=\dfrac{747}{5}-27\\ =\dfrac{612}{5}\)
`@` `\text {Ans}`
`\downarrow`
\(81\cdot\left(-\dfrac{4}{3}\right)^2-25\cdot\left(-\dfrac{3}{5}\right)^3-3^3\)
`=`\(9^2\cdot\left(-\dfrac{4}{3}\right)^2-25\cdot\left(-\dfrac{27}{125}\right)-27\)
`=`\(\left[9\cdot\left(-\dfrac{4}{3}\right)\right]^2-\left(-\dfrac{27}{5}\right)-27\)
`= (-12)^2 + 27/5 - 27`
`= 144 + 27/5 - 27`
`=``612/5`
`@` `\text {Ans}`
`\downarrow`
\(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}?\)
Đặt \(A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)
`3A=`\(3\times\left(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\right)\)
`3A =`\(3+\dfrac{3}{3}+\dfrac{3}{9}+\dfrac{3}{27}+\dfrac{3}{81}+\dfrac{3}{243}+\dfrac{3}{729}\)
`3A =`\(3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)
`3A - A=`\(\left(3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\right)-\left(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\right)\)
`2A =`\(3-\dfrac{1}{729}\)
`2A=`\(\dfrac{2186}{729}\)
\(A=\dfrac{2186}{729}\div2=\dfrac{1093}{729}\)
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