Tìm x biết
a)\(\sqrt{x^2-3}\le x^2-3\)
b) \(\sqrt{x-1}< x+3\)
K
Khách
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18 tháng 1 2022
xem gi
co ban nho cua toi
may bi loi unikey thong cam nha moi nguoi
hihi
18 tháng 1 2022
a) \(\sqrt[3]{x}< 2\Leftrightarrow\left(\sqrt[3]{x}\right)^3< 2^3\Leftrightarrow x< 8\)
b) \(\sqrt[3]{2x-1}>-3\Leftrightarrow\left(\sqrt[3]{2x-1}\right)^3>\left(-3\right)^3\Leftrightarrow2x-1>-27\Leftrightarrow2x>-26\Leftrightarrow x>-13\)
c) \(\sqrt[3]{2-3x}\le1\Leftrightarrow\left(\sqrt[3]{2-3x}\right)^3\le1\Leftrightarrow2-3x\le1\Leftrightarrow3x\ge1\Leftrightarrow x\ge\frac{1}{3}\)
d) \(\sqrt[3]{3-4x}\ge5\Leftrightarrow\left(\sqrt[3]{3-4x}\right)^3\ge5^3\Leftrightarrow3-4x\ge125\Leftrightarrow4x\le-122\Leftrightarrow x\le-\frac{61}{2}\)
9 tháng 7 2018
\(a.x+3+\sqrt{x^2-6x+9}=x+3+\text{ |}x-3\text{ |}=x+3+3-x=6\) \(b.\sqrt{x^2+4x+4}-\sqrt{x^2}=\text{ |}x+2\text{ |}-\text{ |}x\text{ |}=x+2-\left(-x\right)=x+2+x=2x+2\) \(c.\dfrac{\sqrt{x^2-2x+1}}{x-1}=\dfrac{x-1}{x-1}=1\)
\(d.\text{ |}x-2\text{ |}+\dfrac{\sqrt{x^2-4x+4}}{x-2}=\text{ |}x-2\text{ |}+\dfrac{\text{ |}x-2\text{ |}}{x-2}=2-x+\dfrac{-\left(x-2\right)}{x-2}=2-x-1=1-x\)
MN
2 tháng 3 2020
Câu 3 :
\(ĐKXĐ:x>0\)
\(P=\left(\frac{2}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+2}\right):\frac{2\sqrt{x}}{x+2\sqrt{x}}\)
\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}\cdot\frac{x+2\sqrt{x}}{2\sqrt{x}}\)
\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{2\sqrt{x}}\)
b) Để P = 3
\(\Leftrightarrow\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}=3\)
\(\Leftrightarrow2\sqrt{x}+4+x=6\sqrt{x}\)
\(\Leftrightarrow x-4\sqrt{x}+4=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\)
\(\Leftrightarrow\sqrt{x}-2=0\)
\(\Leftrightarrow\sqrt{x}=2\)
\(\Leftrightarrow x=4\)(tm)
Vậy để \(P=3\Leftrightarrow x=4\)
MN
2 tháng 3 2020
Câu 1 : Hình như sai đề !! Mik sửa :
\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
\(A=\left(\frac{x}{x\sqrt{x}-4\sqrt{x}}-\frac{6}{3\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right):\left(\frac{x-4+10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\frac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{6}{\sqrt{x}+2}\)
\(\Leftrightarrow A=\frac{-6\left(\sqrt{x}+2\right)}{6\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(\Leftrightarrow A=-\frac{1}{\sqrt{x}-2}\)
b) Để A < 2
\(\Leftrightarrow-\frac{1}{\sqrt{x}-2}< 2\)
\(\Leftrightarrow-1< 2\sqrt{x}-4\)
\(\Leftrightarrow2\sqrt{x}>3\)
\(\Leftrightarrow\sqrt{x}>1,5\)
\(\Leftrightarrow x>2,25\)
Vậy để \(A< 2\Leftrightarrow x>2,25\)
UK
4 tháng 12 2017
Ta có: \(\left(-x^2+4x+21\right)-\left(-x^2+3x+10\right)=x+11>0\Rightarrow B>0\)
\(B^2=\left(x+3\right)\left(7-x\right)+\left(x+2\right)\left(5-x\right)-2\sqrt{\left(x+3\right)\left(7-x\right)\left(x+2\right)\left(5-x\right)}=\left(\sqrt{\left(x+3\right)\left(5-x\right)}-\sqrt{\left(x+2\right)\left(7-x\right)}\right)^2+2\ge2\)
\(\Rightarrow B\ge\sqrt{2}\)
GTNN của B là \(\sqrt{2}\Leftrightarrow x=\dfrac{1}{3}\)
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 \([0;1)\)
\(f'\left(t\right)=-6t+2=0\Rightarrow t=\dfrac{1}{3}\)
\(f\left(0\right)=0;f\left(\dfrac{1}{3}\right)=\dfrac{1}{3};f\left(1\right)=-1\)
\(\Rightarrow-1< f\left(t\right)\le\dfrac{1}{3}\)
\(\Rightarrow-1< m\le\dfrac{1}{3}\)
AM
26 tháng 12 2021
\(A=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)\(A=\dfrac{2x-6\sqrt{x}+x+\sqrt{x+}3\sqrt{x}+3+3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)\(A=\dfrac{3x-13\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
28 tháng 3 2019
\(f\left(x;y\right)=x+y+x\sqrt{1-y^2}+y\sqrt{1-x^2}\)
\(\Rightarrow\frac{\sqrt{3}}{2}f\left(x;y\right)=\frac{\sqrt{3}}{2}\left(x+y\right)+\frac{1}{2}\left(x\sqrt{3-3y^2}+y\sqrt{3-3x^2}\right)\)
\(\Rightarrow\frac{\sqrt{3}}{2}f\left(x;y\right)\le\frac{\frac{3}{4}+x^2+\frac{3}{4}+y^2}{2}+\frac{1}{2}\left(\frac{-3x^2+y^2+3-3y^2+x^2+3}{2}\right)\)
\(\Rightarrow\frac{\sqrt{3}}{2}f\left(x;y\right)\le\frac{\frac{3}{2}+x^2+y^2-x^2-y^2+3}{2}=\frac{9}{4}\)
\(\Rightarrow f\left(x;y\right)\le\frac{3\sqrt{3}}{2}\)
Dấu "=" khi x = y = \(\frac{\sqrt{3}}{2}\).
#Kaito#