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AH
Akai Haruma
Giáo viên
16 tháng 12 2021
Ý bạn là \(f(x)=\frac{1}{9+x^2}+\frac{3}{9+x^2}\) hay thế nào? Bạn cần viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo để được hỗ trợ tốt hơn).
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 8 2021
ĐKXĐ: \(x\ge1\)
\(3\sqrt[]{x-1}+m\sqrt[]{x+1}=2\sqrt[4]{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow3\sqrt[]{\dfrac{x-1}{x+1}}+m=2\sqrt[4]{\dfrac{x-1}{x+1}}\)
Đặt \(\sqrt[4]{\dfrac{x-1}{x+1}}=t\Rightarrow0\le t< 1\)
\(\Rightarrow3t^2+m=2t\Leftrightarrow-3t^2+2t=m\)
Xét \(f\left(t\right)=-3t^2+2t\) trên \([0;1)\)
\(f'\left(t\right)=-6t+2=0\Rightarrow t=\dfrac{1}{3}\)
\(f\left(0\right)=0;f\left(\dfrac{1}{3}\right)=\dfrac{1}{3};f\left(1\right)=-1\)
\(\Rightarrow-1< f\left(t\right)\le\dfrac{1}{3}\)
\(\Rightarrow-1< m\le\dfrac{1}{3}\)
OG
2
22 tháng 6 2020
Xét: \(1+\frac{1}{x\left(x+2\right)}=\frac{x\left(x+2\right)+1}{x\left(x+2\right)}=\frac{x^2+2x+1}{x\left(x+2\right)}=\frac{\left(x+1\right)^2}{x\left(x+2\right)}\)
Viết A = \(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{17.19}\right)\left(1+\frac{1}{18.20}\right)\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{16^2}{17.19}\frac{19^2}{18.20}\)
\(=\frac{2}{1}.\frac{19}{20}=\frac{19}{10}\)
H9
HT.Phong (9A5)
CTVHS
4 tháng 9 2023
\(A=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{9x-1}\right):\left(1-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\) (ĐK: \(x\ge0;x\ne\dfrac{1}{9}\))
\(A=\left[\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}\right)^2-1^2}\right]:\left[\dfrac{\left(3\sqrt{x}+1\right)\cdot1}{3\sqrt{x}+1}-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right]\)
\(A=\left[\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(A=\left[\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}-\dfrac{3\sqrt{x}-1}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\dfrac{3}{3\sqrt{x}+1}\)
\(A=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\cdot\dfrac{3\sqrt{x}+1}{3}\)
\(A=\dfrac{3x+3\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\cdot\dfrac{3\sqrt{x}+1}{3}\)
\(A=\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\cdot\dfrac{3\sqrt{x}+1}{3}\)
\(A=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{3\sqrt{x}-1}\)
\(A=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\)
4 tháng 9 2023
\(A=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right):\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\cdot\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1+5\sqrt{x}+1}{3\sqrt{x}-1}\cdot\dfrac{1}{3}\)
\(=\dfrac{3x+3\sqrt{x}}{3\sqrt{x}-1}\cdot\dfrac{1}{3}\)
\(=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\)
QD
31 tháng 3 2017
a) =
=
b) =
=
=
. ( Với điều kiện b # 1)
c) \(\dfrac{a^{\dfrac{1}{3}}b^{-\dfrac{1}{3}-}a^{-\dfrac{1}{3}}b^{\dfrac{1}{3}}}{\sqrt[3]{a^2}-\sqrt[3]{b^2}}\)= =
=
( với điều kiện a#b).
d) \(\dfrac{a^{\dfrac{1}{3}}\sqrt{b}+b^{\dfrac{1}{3}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}\) = =
=
=
25 tháng 11 2023
a: \(y=\left(2x^2-x+1\right)^{\dfrac{1}{3}}\)
=>\(y'=\dfrac{1}{3}\left(2x^2-x+1\right)^{\dfrac{1}{3}-1}\cdot\left(2x^2-x+1\right)'\)
\(=\dfrac{1}{3}\cdot\left(4x-1\right)\left(2x^2-x+1\right)^{-\dfrac{2}{3}}\)
b: \(y=\left(3x+1\right)^{\Omega}\)
=>\(y'=\Omega\cdot\left(3x+1\right)'\cdot\left(3x+1\right)^{\Omega-1}\)
=>\(y'=3\Omega\left(3x+1\right)^{\Omega-1}\)
c: \(y=\sqrt[3]{\dfrac{1}{x-1}}\)
=>\(y'=\dfrac{\left(\dfrac{1}{x-1}\right)'}{3\cdot\sqrt[3]{\left(\dfrac{1}{x-1}\right)^2}}\)
\(=\dfrac{\dfrac{1'\left(x-1\right)-\left(x-1\right)'\cdot1}{\left(x-1\right)^2}}{\dfrac{3}{\sqrt[3]{\left(x-1\right)^2}}}\)
\(=\dfrac{-x}{\left(x-1\right)^2}\cdot\dfrac{\sqrt[3]{\left(x-1\right)^2}}{3}\)
\(=\dfrac{-x}{\sqrt[3]{\left(x-1\right)^4}\cdot3}\)
d: \(y=log_3\left(\dfrac{x+1}{x-1}\right)\)
\(\Leftrightarrow y'=\dfrac{\left(\dfrac{x+1}{x-1}\right)'}{\dfrac{x+1}{x-1}\cdot ln3}\)
\(\Leftrightarrow y'=\dfrac{\left(x+1\right)'\left(x-1\right)-\left(x+1\right)\left(x-1\right)'}{\left(x-1\right)^2}:\dfrac{ln3\left(x+1\right)}{x-1}\)
\(\Leftrightarrow y'=\dfrac{x-1-x-1}{\left(x-1\right)^2}\cdot\dfrac{x-1}{ln3\cdot\left(x+1\right)}\)
\(\Leftrightarrow y'=\dfrac{-2}{\left(x-1\right)\cdot\left(x+1\right)\cdot ln3}\)
e: \(y=3^{x^2}\)
=>\(y'=\left(x^2\right)'\cdot ln3\cdot3^{x^2}=2x\cdot ln3\cdot3^{x^2}\)
f: \(y=\left(\dfrac{1}{2}\right)^{x^2-1}\)
=>\(y'=\left(x^2-1\right)'\cdot ln\left(\dfrac{1}{2}\right)\cdot\left(\dfrac{1}{2}\right)^{x^2-1}=2x\cdot ln\left(\dfrac{1}{2}\right)\cdot\left(\dfrac{1}{2}\right)^{x^2-1}\)
h: \(y=\left(x+1\right)\cdot e^{cosx}\)
=>\(y'=\left(x+1\right)'\cdot e^{cosx}+\left(x+1\right)\cdot\left(e^{cosx}\right)'\)
=>\(y'=e^{cosx}+\left(x+1\right)\cdot\left(cosx\right)'\cdot e^u\)
\(=e^{cosx}+\left(x+1\right)\cdot\left(-sinx\right)\cdot e^u\)
ND
Nguyễn Đức Trí
VIP
25 tháng 11 2023
a) \(y=\left(2x^2-x+1\right)^{\dfrac{1}{3}}\)
\(\Rightarrow y'=\dfrac{1}{3}.\left(2x^2-x+1\right)^{\dfrac{1}{3}-1}.\left(4x-1\right)\)
\(\Rightarrow y'=\dfrac{1}{3}.\left(2x^2-x+1\right)^{-\dfrac{2}{3}}.\left(4x-1\right)\)
b) \(y=\left(3x+1\right)^{\pi}\)
\(\Rightarrow y'=\pi.\left(3x+1\right)^{\pi-1}.3=3\pi.\left(3x+1\right)^{\pi-1}\)
c) \(y=\sqrt[3]{\dfrac{1}{x-1}}\)
\(\Rightarrow y'=\dfrac{\left(x-1\right)^{-1-1}}{3\sqrt[3]{\left(\dfrac{1}{x-1}\right)^{3-1}}}=\dfrac{\left(x-1\right)^{-2}}{3\sqrt[3]{\left(\dfrac{1}{x-1}\right)^2}}=\dfrac{1}{3.\sqrt[]{x-1}.\sqrt[3]{\left(\dfrac{1}{x-1}\right)^2}}\)
\(\Rightarrow y'=\dfrac{1}{3\left(x-1\right)^{\dfrac{1}{2}}.\left(x-1\right)^{\dfrac{2}{3}}}=\dfrac{1}{3\left(x-1\right)^{\dfrac{7}{6}}}=\dfrac{1}{3\sqrt[6]{\left(x-1\right)^7}}\)
d) \(y=\log_3\left(\dfrac{x+1}{x-1}\right)\)
\(\Rightarrow y'=\dfrac{\dfrac{1-\left(-1\right)}{\left(x-1\right)^2}}{\dfrac{x+1}{x-1}.\ln3}=\dfrac{2}{\left(x+1\right)\left(x-1\right).\ln3}\)
e) \(y=3^{x^2}\)
\(\Rightarrow y'=3^{x^2}.ln3.2x=2x.3^{x^2}.ln3\)
f) \(y=\left(\dfrac{1}{2}\right)^{x^2-1}\)
\(\Rightarrow y'=\left(\dfrac{1}{2}\right)^{x^2-1}.ln\dfrac{1}{2}.2x=2x.\left(\dfrac{1}{2}\right)^{x^2-1}.ln\dfrac{1}{2}\)
Các bài còn lại bạn tự làm nhé!
NH
1
AH
Akai Haruma
Giáo viên
30 tháng 7 2021
Lời giải:
$y'=x^2-(m-1)x-m=(x+1)(x-m)$
$y''=2x-(m-1)$
Nếu $x_{ct}=-1$ thì $y''(-1)=-1-m>0\Leftrightarrow m< -1$
$y_{ct}=\frac{1}{2}m+\frac{1}{2}=\frac{1}{3}$
$\Leftrightarrow m=\frac{-1}{3}$ (loại vì $m< -1$)
Nếu $x_{ct}=m$ thì $y''(m)=m+1>0\Leftrightarrow m>-1$
$y_{ct}=\frac{-1}{6}m^3+\frac{1}{2}m^2+\frac{1}{3}=\frac{1}{3}$
$\Leftrightarrow m=0$ (chọn) hoặc $m=-3$ (loại)
Vậy $m=0$
1 + 1 = 2 ( từ lớp 12 đá 1 phát xuống lớp gấu bông học lại đê :) )
1 + 1 = 2 CHỨ SAO LẠI BẰNG 3