Rút gọn biểu thức:
A = \([(32)^{\dfrac{2}{3}}]^{\dfrac{-2}{5}}\)
B= \(\dfrac{x^{-2}+y^{-2}}{x^{-1}+y^{-1}}\)
C = \((a^{\dfrac{1}{3}}-b^{\dfrac{2}{3}})(a^{\dfrac{2}{3}}+a^{\dfrac{1}{3}}×b^{\dfrac{4}{3}}+b^{\dfrac{4}{9}})\)
D = \((x+y^\dfrac{3}{2}÷\sqrt{x})^\dfrac{2}{3}÷[\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}}+\dfrac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}]^\dfrac{2}{3}\)
E = \([\dfrac{1}{x^{\dfrac{1}{2}}-4x^{\dfrac{-1}{2}}}-\dfrac{2\sqrt[3]{x}}{x\sqrt[3]{x}-4\sqrt[3]{x}}]^{-2}-\sqrt{x^2+8x+16}\)
giúp mình mấy bài này với ạ.
\(4^{2x+\sqrt{x+2}}+2^{x^3}=4^{2+\sqrt{x+2}}+2^{x^3+4x-4}\)
\(4^{\sqrt[3]{x+5}+1}+2.2^{\sqrt[3]{x+5}+x}=2.4^x\)
\(5^{\dfrac{1}{2}}+5^{\dfrac{1}{2}+log_5sinx}=15^{\dfrac{1}{2}log_{15}cosx}\)
\(2^{log_{\sqrt{3}}\left|x+1\right|}.5^{log_3\left|x+1\right|}< 400\)
Tính đạo hàm của các hàm số sau:
a) \(y = (2x^2 - x + 1)^{\frac{1}{3}}\)
b) \(y = (3x+1)^{\pi}\)
c) \(y = \sqrt[3]{\dfrac{1}{x-1}}\)
d) \(y =\log_{3} \left(\dfrac{x+1}{x-1}\right)\)
e) \(y = 3^{x^{2}}\)
f) \(y = \left(\dfrac{1}{2}\right)^{x^2-1}\)
h) \(y = (x+1) . e^{cosx}\)
g) \(y = \ln (x^2+x+1)\)
l) \(y = \dfrac{\ln x}{x+1}\)
Giải phương trình:
a) \(2log_2x+log_{\dfrac{1}{2}}\left(1-\sqrt{x}\right)=\dfrac{1}{2}log_{\sqrt{2}}\left(x-2\sqrt{x}+2\right)\)
b) \(log_3\dfrac{x^2-2x+1}{x}+x^2+1=3x\)
Giúp mình hai câu này với ạ.
1. Tìm tập xác định của các hàm số sau:
a) \(y = 3(x-1)^{-3}\)
b) \(y = (2 - x^2)^{\frac{2}{5}}\)
c) \(y = (x^2 + x - 6)^{\frac{-1}{3}}\)
d) \(y = \left(\dfrac{1}{x^2-1}\right)^3\)
e) \(y = \log_{3} (x^2-2)\)
f) \(y = \log_{\frac{1}{2}}\sqrt{x-1}\)
g) \(y = \log_{\pi} (x^2+x-6)\)
1.xét sự biến thiên của hàm số
y=\(-x^4+\dfrac{4}{3}x^3+1\)
y=\(-x^4-2x^2+3\)
y=\(\dfrac{x^2-2x+2}{x-1}\)
y=\(\sqrt{16-^{ }x^2}\)
2.tìm m để hàm số nghịch biến trên TXĐ
y=\(x^3+mx^2+mx+1\)
y=\(\dfrac{mx+1}{x+m}\)
y=\(mx^2+2mx+m\)
y=\(\dfrac{mx^2-2x+1}{x-2}\)
tính P
P=\(\sqrt[4]{\dfrac{3+2\sqrt{2}}{3-2\sqrt{2}}}-\sqrt[4]{\dfrac{3-2\sqrt{2}}{3+2\sqrt{2}}}\)
Giải phương trình
\(\sqrt{\dfrac{1-x}{x}}=\dfrac{2x+x^2}{1+x^2}\)
Bài 1:
a)
ta có: \(\dfrac{50}{100}=\dfrac{1}{2};\dfrac{-\dfrac{4}{13}}{-\dfrac{8}{13}}=\dfrac{1}{2};\dfrac{\dfrac{2}{15}}{\dfrac{4}{15}}=\dfrac{1}{2};\dfrac{-\dfrac{2}{17}}{-\dfrac{4}{17}}=\dfrac{1}{2}\)
\(\dfrac{50}{100}=\dfrac{\dfrac{4}{13}}{\dfrac{8}{13}}=\dfrac{\dfrac{2}{15}}{\dfrac{4}{15}}=\dfrac{\dfrac{2}{17}}{\dfrac{4}{17}}=\dfrac{50-\dfrac{4}{13}+\dfrac{2}{15}-\dfrac{2}{17}}{100-\dfrac{8}{13}+\dfrac{4}{15}-\dfrac{4}{17}}=\dfrac{1}{2}\)
vậy \(A=\dfrac{1}{2}\)
b)
\(B=\dfrac{1}{19}+\dfrac{9}{19.29}+\dfrac{9}{29.39}+...+\dfrac{9}{1999.2009}\\ B=\dfrac{1}{19}-\dfrac{1}{19}+\dfrac{2}{29}-\dfrac{2}{29}+\dfrac{3}{39}-...-\dfrac{199}{1999}+\dfrac{200}{2009}\\ B=\dfrac{200}{2009}\)
Bài 2:
\(\dfrac{a}{b}=\dfrac{b}{3c}=\dfrac{c}{9a}=\dfrac{b+c}{3c+9a}\)
suy ra: \(b=\dfrac{3c\left(b+c\right)}{3c+9a}=\dfrac{3cb+3c^2}{3c+9a}=\dfrac{bc+c^2}{c+3a}\)
\(c=\dfrac{9a\left(b+c\right)}{3c+9a}=\dfrac{9ab+9ac}{3c+9a}=\dfrac{3ab+3ac}{c+3a}\)
giả sử b=c là đúng thì :\(\dfrac{bc+c^2}{c+3a}=\dfrac{3ab+3ac}{c+3a}\)
hay \(bc+c^2=3ab+3ac\\ \Leftrightarrow c^2+bc-3ab-3ac=0\)
\(\Leftrightarrow\left(b+c\right)\left(c-3a\right)=0\Rightarrow c-3a=0\Rightarrow c=3a\)
b) \(\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{2013.2015}+\dfrac{1}{2014.2016}\\ =\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{2.4}+\dfrac{2}{3.5}+...+\dfrac{2}{2013.2015}+\dfrac{2}{2014.2016}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{2016}\right)=\dfrac{2015}{4032}< 1\)
mà \(1< \dfrac{4}{3}\) nên \(\dfrac{2015}{4032}< \dfrac{4}{3}\)
hay \(\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{2013.2015}+\dfrac{1}{2014.2016}< \dfrac{4}{3}\)
bài 3:
a)\(\left(x-y\right)\left(x+y\right)=x^2-y^2-xy+xy=x^2-y^2\) (đpcm)
b) áp dụng BĐT tam giác, ta có:
\(a+b>c\Rightarrow a+b-c>0\\ b+c>a\Rightarrow b+c-a< 0\\ a+c>b\Rightarrow a-b+c>0\)
suy ra: \(\left(a+b-c\right)\left(b+c-a\right)\left(a-b+c\right)< 0\: \: \: \: \: \: \)
đồng thời \(abc>0\) với mọi a, b, c dương.
nên \(\left(a+b-c\right)\left(b+c-a\right)\left(a-b+c\right)< abc\)
ko tìm dc dấu bằng xảy ra.
1)Cho X = log\(\dfrac{1}{2}\)+log\(\dfrac{2}{3}\)+...+log\(\dfrac{99}{100}\). Chọn câu trả lời đúng về giá trị của X:
a)X>2 b)X=0 c) X=-2 d)X=\(\dfrac{1}{2}\)
2)Đặt log32 =a ,log35 = b. X=log3\(\dfrac{1}{2}\)+log3\(\dfrac{2}{3}\)+....+log3\(\dfrac{99}{100}\). X được biểu thị qua a,b là:
a) X=-2a-2b b)X =-2a+2b
c)X =2a-2b c)X =2a+2b
