Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1: \(\frac{a^2+c^2}{b^2+c^2}=\frac{a}{b}\) (1)
Từ \(\frac{a}{c}=\frac{c}{b}\Rightarrow ab=c^2\)
Thay vào (1) ta có:
\(\frac{a^2+ab}{b^2+ab}=\frac{a}{b}\Rightarrow\frac{a\left(a+b\right)}{b\left(a+b\right)}=\frac{a}{b}\) (luôn đúng)
Vậy ta có điều phải chứng minh
\(=\left(x^2+2x+1\right)+\left(y^2-8y+16\right)=\left(x+1\right)^2+\left(y-4\right)^2\ge0\forall x,y\)
dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=4\end{matrix}\right.\)
Bài 1:
a) \(2x\left(x^2-5x+6\right)=2x^3-10x^2+12x\)
b) \(\left(7x^5+14x^2y^3-28x^3y^2\right):7x^2=x^3+2y^3-4xy^2\)
Bài 2:
\(x^2+y^2+2x-8y+17=\left(x^2+2x+1\right)+\left(y^2-8y+16\right)=\left(x+1\right)^2+\left(y-4\right)^2\ge0\forall x,y\)
a: Ta có: \(\left(8\cdot5^7+5^6-5^5\right):5^5\)
\(=8\cdot5^2+5-1\)
\(=200+4=204\)
b: Ta có: \(\left(9^{30}-27^{19}\right):3^{57}+\left(125^9-25^{12}\right):5^{24}\)
\(=3^{60}:3^{57}-3^{57}:3^{57}+5^{27}:5^{24}-5^{24}:5^{24}\)
\(=27-1+125-1\)
=150
a. (8,57 - 55 + 56) : 55
= (8,57 : 55) - (55 : 55) + (56 : 55)
= 1,72 - 1 + 5
= 2,89 - 1 + 5
= 6,89
b. (930 - 2719) : 357 + (1259 - 2512) : 524
= (930 : 357) - (2719 : 357) + (1259 : 524) - (2512 : 524)
= 33 - 1 + 125 - 1
= 27 - 1 + 125 - 1
= 150
c. (1012 + 511 . 29 - 513 - 28) : 4 . 55 . 106
= (1012 + 2,5 , 1010 - 513 - 28) : 1,25 . 1010
= (1012 : 1,25 . 1010) + (2,5 . 1010 : 1,25 . 1010) - (513 : 1,25 . 1010) - (28 : 1,25 . 1010)
= 80 + 2 - \(\dfrac{25}{256}\) - \(\dfrac{1}{48828125}\)
= 81,90234373 \(\approx\) 82
2:
a: =>x-1=0 hoặc 3x+1=0
=>x=1 hoặc x=-1/3
b: =>x-5=0 hoặc 7-x=0
=>x=5 hoặc x=7
c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)
d: =>x=0 hoặc x^2-1=0
=>\(x\in\left\{0;1;-1\right\}\)
a,\(\dfrac{3-x}{x-5}+\dfrac{2x-8}{x-5}=\dfrac{3-x+2x-8}{x-5}=\dfrac{x-5}{x-5}=1\)
b, \(\dfrac{1}{x-y}+\dfrac{1}{x+y}+\dfrac{2x}{x^2-y^2}=\dfrac{x+y}{\left(x-y\right)\left(x+y\right)}+\dfrac{x-y}{\left(x-y\right)\left(x+y\right)}+\dfrac{2x}{\left(x-y\right)\left(x+y\right)}=\dfrac{x+y+x-y+2x}{\left(x-y\right)\left(x+y\right)}=\dfrac{4x}{\left(x-y\right)\left(x+y\right)}\)
a) 81^11.3^17/27^10.9^15
=(9^2)^11.3^17/(3^3)^10.9^15
=3^44.3^17/3^30.3^30
=3^61/3^60
=3
b) A =( (2^12.3^5 - 2^12.3^4)/ (2^12.3^6 + 2^12.3^5) ) - ((5^10.7^3 - 5^10.7^4)/(5^9.7^3 + 5^9.2^3.7^3))
=(2^12.3^4(3-1))/2^12.3^5(3+1) - 5^10.7^3(1-7)/5^9.7^3(1+8)
=2/12- (-30/9)=1/6 + 10/3 = 7/2
3/6, /8,4/6