1/ Tìm giá trị nhỏ nhất của biểu thức: A= |x-20| + |y+5| - 2015
2/ Tính: \(B=\left(1+\frac{7}{9}\right)\left(1+\frac{7}{20}\right)\left(1+\frac{7}{33}\right)....\left(1+\frac{7}{2900}\right)\)
tính tổng\(A=\left(1+\frac{7}{9}\right).\left(1+\frac{7}{20}\right).\left(1+\frac{7}{33}\right)....\left(1+\frac{7}{2900}\right)\)
\(A=\left(1+\frac{7}{9}\right)\left(1+\frac{7}{20}\right)\left(1+\frac{7}{33}\right)..........\left(1+\frac{7}{2900}\right)\)
\(A=\frac{16}{9}.\frac{27}{20}.\frac{40}{33}.......\frac{2907}{2900}\)
\(A=\frac{2.8}{1.9}.\frac{3.9}{2.10}.\frac{4.10}{3.11}......\frac{51.57}{50.58}\)
\(A=\frac{2.3.4.....51}{1.2.3...50}.\frac{8.9.10....57}{9.10.11...58}\)
\(A=51.\frac{8}{58}=\frac{204}{29}\)
Bạn Nguyễn Tuấn Minh làm đúng rùi đó !!! Chuẩn ý kiến mk...^.^
Tính
\(A=\left(1+\frac{7}{9}\right)\left(1+\frac{7}{20}\right)\left(1+\frac{7}{33}\right)....\left(1+\frac{7}{2900}\right)\)
Cô giải như sau Minh nhé :)
\(A=\left(1+\frac{7}{9}\right)\left(1+\frac{7}{20}\right)\left(1+\frac{7}{33}\right)...\left(1+\frac{7}{2900}\right)=\frac{16}{9}.\frac{27}{20}.\frac{40}{33}...\frac{2907}{2900}\)
\(=\frac{8.2}{9.1}.\frac{9.3}{10.2}.\frac{10.4}{11.3}....\frac{57.51}{58.50}=\frac{\left(8.9.10....57\right)\left(2.3.4...51\right)}{\left(9.10.11...58\right)\left(2.3.4....50\right)}=\frac{8.51}{58}=\frac{204}{29}\)
( 1 + 7/9 ) x ( 1 + 7/20 ) x ( 1 + 7/33 ) x...x ( 1 + 7/2900)
= (8x2)/(9x1) x (9x3)/(10x2) x (10x4)/(11x3) x...x (57x51)(58x50)
=(8x2x9x3x10x4x...x57x51) / (9x1x10x2x11x3x...x58x50) Sau khi giản ước ta được :
= (8x51) / (1x58) = 204/29
tính \(B=\left(1+\frac{7}{9}\right)\left(1+\frac{7}{20}\right)\left(1+\frac{7}{33}\right)...\left(1+\frac{7}{2900}\right)\)
Nhanh giùm mình nha, gấp lắm rùi
\(H=\left(1+\frac{7}{9}\right)x\left(1+\frac{7}{20}\right)x\left(1+\frac{7}{33}\right)x....x\left(1+\frac{7}{2900}\right)\)
Tìm H giúp mình với mấy bạn, nhớ là làm theo cách làm của cấp 1 nhé!
Chứng minh rằng:
\(\left(1+\frac{7}{9}\right).\left(1+\frac{7}{20}\right).\left(1+\frac{7}{33}\right)....\left(1+\frac{7}{2900}\right)=7\frac{1}{29}\)
CMR : \(\left(1+\frac{7}{7}\right)+\left(1+\frac{7}{20}\right)+\left(1+\frac{7}{33}\right)+...+\left(1+\frac{7}{2900}\right)=7\frac{1}{29}\)
Bài 1 : Tính
\(a,\left(\frac{1^{ }}{2}\right)^{15}\cdot\left(\frac{1}{4}\right)^{20}\)
b, \(\left(\frac{1}{9}\right)^{25}:\left(\frac{1}{3}\right)^{30}\)
Bài 2 : Chứng minh rằng 7^6 + 7^5 - 7^4 chia hết cho 55
Tính A = 1+5+5^2 + 5^3 +...+5^49+5^50
bÀI 3
a, tìm giá trị lớn nhất của biểu thức :
\(A=\frac{3}{\left(X+2\right)^2+4}\)
B, tÌM GIÁ trị nhỏ nhất :
\(B=\left(x+1\right)^2+\left(y+3\right)^2+1\)
Bài 4 : Chứng minh rằng góc tạo bỏi 2 tia phân giác của 2 góc kề bù là góc vuông
Bài 4:
Giải:
Vì Om là tia phân giác của góc xOz nên:
mOz = 1/2.xOz
Vì On là tia phân giác của góc zOy nên:
zOn = 1/2 . zOy
Ta có: xOz + zOy = 180o ( kề bù )
=> 1/2(xOz + zOy) = 1/2 . 180o
=> 1/2.xOz + 1/2.zOy = 90o
=> mOz + zOn = 90o
=> mOn = 90o (đpcm)
Bài 2:
7^6 + 7^5 - 7^4 = 7^4.( 7^2 + 7 - 1 ) = 7^4 . 55 chia hết cho 55
Vậy 7^6 + 7^5 - 7^4 chia hết cho 55
A = 1 + 5 + 5^2 + ... + 5^50
=> 5A = 5 + 5^2 + 5^3 + ... + 5^51
=> 5A - A = ( 5 + 5^2 + 5^3 + ... + 5^51 ) - ( 1 + 5 + 5^2 + ... + 5^50 )
=> 4A = 5^51 - 1
=> A = ( 5^51 - 1 )/4
Tính nhanh giá trị biểu thức sau :
A= \(\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)+\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+9\right)}+\frac{1}{\left(x+9\right)\left(x+11\right)}\)
\(A=\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+9\right)}+\frac{1}{\left(x+9\right)\left(x+11\right)}\)
\(=\frac{1+1+1+1+1}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)\left(x+9\right)\left(x+11\right)}\)
\(=\frac{5}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)\left(x+9\right)\left(x+11\right)}\)
\(=\frac{5}{\left(x+1\right)\left(x+11\right)\left(x+3\right)\left(x+9\right)\left(x+5\right)\left(x+7\right)}\)
\(=\frac{5}{\left(x^2+11x+x+11\right)\left(x^2+9x+3x+27\right)\left(x^2+7x+5x+35\right)}\)
\(=\frac{5}{\left(x^2+12x+11\right)\left(x^2+12x+27\right)\left(x^2+12x+35\right)}\)
A=\(\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+7}+\frac{1}{x+7}-\frac{1}{x+9}+\frac{1}{x+9}-\frac{1}{x+11}\)
Rút gọn hết đi ta có \(\frac{1}{x+1}-\frac{1}{x+11}\)=\(\frac{x+11}{\left(x+1\right).\left(x+11\right)}-\frac{x+1}{\left(x+1\right).\left(x+11\right)}\)
A=\(\frac{x+11-x-1}{\left(x+1\right).\left(x+11\right)}\)
A=\(\frac{10}{x^2+12x+11}\)
\(A=\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+9\right)}+\frac{1}{\left(x+9\right)\left(x+11\right)}\)
\(A=\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+...+\frac{1}{x+9}-\frac{1}{x+11}\)
\(A=\frac{1}{x+1}-\frac{1}{x+11}\Leftrightarrow A=\frac{x+11}{\left(x+1\right)\left(x+11\right)}+\frac{x+1}{\left(x+1\right)\left(x+11\right)}\)
\(\Leftrightarrow A=\frac{2x+12}{\left(x+1\right)\left(x+11\right)}\)