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NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 3 2019
a/ \(\left(x+3\right)\left(3\left(x^2+1\right)^2+2\left(x+3\right)^2\right)=5\left(x^2+1\right)^3\)
\(\Leftrightarrow3\left(x+3\right)\left(x^2+1\right)^2+2\left(x+3\right)^3-5\left(x^2+1\right)^3=0\)
\(\Leftrightarrow3\left(x+3\right)\left(x^2+1\right)^2-3\left(x^2+1\right)^3+2\left(x+3\right)^3-2\left(x^2+1\right)^3=0\)
\(\Leftrightarrow3\left(x^2+1\right)^2\left(-x^2+x+2\right)+2\left(-x^2+x+2\right)\left(\left(x+3\right)^2+\left(x+3\right)\left(x^2+1\right)+\left(x^2+1\right)^2\right)=0\)
\(\Leftrightarrow\left(-x^2+x+2\right)\left[3\left(x^2+1\right)^2+2\left(x+3+\dfrac{x^2+1}{2}\right)^2+\dfrac{3\left(x^2+1\right)^2}{4}\right]=0\)
\(\Leftrightarrow-x^2+x+2=0\) (phần ngoặc phía sau luôn dương)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 3 2019
b/ \(3\left(x^2+2x-1\right)^2-2\left(x^2+3x-1\right)^2+5\left(x^2+3x-1-\left(x^2+2x-1\right)\right)^2=0\)
Đặt \(\left\{{}\begin{matrix}a=x^2+2x-1\\b=x^2+3x-1\end{matrix}\right.\)
\(3a^2-2b^2+5\left(b-a\right)^2=0\Leftrightarrow8a^2+3b^2-10ab=0\)
\(\Leftrightarrow\left(4a-3b\right)\left(2a-b\right)=0\Leftrightarrow\left[{}\begin{matrix}4a=3b\\2a=b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4\left(x^2+2x-1\right)=3\left(x^2+3x-1\right)\\2\left(x^2+2x-1\right)=x^2+3x-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-x-1=0\\x^2+x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=\dfrac{1-\sqrt{5}}{2}\\x=\dfrac{-1+\sqrt{5}}{2}\\x=\dfrac{-1-\sqrt{5}}{2}\end{matrix}\right.\)
N
25 tháng 2 2019
Có: \(x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)^3-3x^3.\dfrac{1}{x^3}\left(x+\dfrac{1}{x}\right)\)\(=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)\)
Có: \(x^6+\dfrac{1}{x^6}=\left(x^2+\dfrac{1}{x^2}\right)^3-3\left(x^2+\dfrac{1}{x^2}\right)\)\(=\left[\left(x+\dfrac{1}{x}\right)^2-2\right]^3-3\left[\left(x+\dfrac{1}{x}\right)^2-2\right]\)
Đặt \(a=x+\dfrac{1}{x}\left(a\ge2\right)\)
\(P=\dfrac{a^6-\left[a^2-2\right]^3+3a^2+4}{a^3+a^3-3a}\)
\(P=\dfrac{-6a^4+15a^2+4}{2a^3-3a}\)
\(\Rightarrow6a^4+2Pa^3-15a^2-3Pa-4=0\)
\(\Rightarrow a^2\left(6a^2+2P+14\right)-\left(14a^2+3Pa+4\right)=0\)
Để pt \(\left\{{}\begin{matrix}6a^2+2P+14\\14a^2+3Pa+4\end{matrix}\right.\) có nghiệm thì
\(\left\{{}\begin{matrix}4P^2-336\ge0\\9P^2-224\ge0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}P\le-2\sqrt{21}\\P\ge2\sqrt{21}\end{matrix}\right.\\\left[{}\begin{matrix}P\le-\dfrac{4\sqrt{14}}{3}\\P\ge\dfrac{4\sqrt{14}}{3}\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow P_{min}=\dfrac{4\sqrt{14}}{3}\)
30 tháng 6 2021
c) Ta có: \(C=\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{\sqrt{x}}{x-4}\)
\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)
d)
Sửa đề: \(D=\dfrac{8+x\left(1+\sqrt{x-2\sqrt{x}+1}\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{x-3\sqrt{x}}{2\left(x-\sqrt{x}-6\right)}\)
Ta có: \(D=\dfrac{8+x\left(1+\sqrt{x-2\sqrt{x}+1}\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{x-3\sqrt{x}}{2\left(x-\sqrt{x}-6\right)}\)
\(=\dfrac{8+x\left(1+\sqrt{x}-1\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x\sqrt{x}+8}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{1}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2\left(\sqrt{x}+2\right)+\sqrt{x}\left(\sqrt{x}-2\right)}{2\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2\sqrt{x}+4+x-2\sqrt{x}}{2\left(x-4\right)}\)
\(=\dfrac{x+4}{2x-8}\)
18 tháng 4 2023
1:
\(=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{2}{3\sqrt{x}-6}\right):\dfrac{2\sqrt{x}+3}{3\sqrt{x}}\)
\(=\dfrac{3+2\sqrt{x}}{3\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{3\sqrt{x}}{2\sqrt{x}+3}=\dfrac{1}{\sqrt{x}-2}\)
AT
30 tháng 6 2021
a) ĐKXĐ có thêm \(x\ne4\)
\(A=\left(\dfrac{x-\sqrt{x}+2}{x-\sqrt{x}-2}-\dfrac{x}{x-2\sqrt{x}}\right):\dfrac{1-\sqrt{x}}{2-\sqrt{x}}\)
\(=\left(\dfrac{x-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}\right).\dfrac{2-\sqrt{x}}{1-\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(x-\sqrt{x}+2\right)-x\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\dfrac{2-\sqrt{x}}{1-\sqrt{x}}\)
\(=\dfrac{-2x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\dfrac{-2\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}-2}{\sqrt{x}-1}=\dfrac{-2}{\sqrt{x}+1}\)
\(B=\left(\dfrac{x}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\dfrac{x+1}{\sqrt{x}+3}:\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{x+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+1}{\sqrt{x}+3}:\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{x+1}{\sqrt{x}+3}.\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+1}{\sqrt{x}+1}\)
5 tháng 8 2022
Bài 2:
a: \(A=\left|5x+1\right|-\dfrac{3}{8}>=-\dfrac{3}{8}\)
Dấu '=' xảy ra khi x=-1/5
b: \(B=\left|-\dfrac{1}{6}x+2\right|+0.25>=0.25\)
Dấu '=' xảy ra khi x=12
Bài 3:
a: \(A=2018-\left|x+2019\right|< =2018\)
Dấu '=' xảy ra khi x=-2019
b: \(=-10-\left|2x-\dfrac{1}{1009}\right|< =-10\)
Dấu '=' xảy ra khi x=1/2018
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
27 tháng 1
\(Đặt:\left\{{}\begin{matrix}a=\dfrac{1}{\left|x+3\right|}\left(ĐK:x\ne-3\right)\\b=\dfrac{1}{\left|y\right|-2}\left(ĐK:y\ne\pm2\right)\end{matrix}\right.\\ Có:\left\{{}\begin{matrix}\dfrac{1}{\left|x+3\right|}+\dfrac{4}{\left|y\right|-2}=\dfrac{11}{6}\\\dfrac{5}{\left|x+3\right|}+\dfrac{2}{\left|y\right|-2}=\dfrac{11}{6}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+4b=\dfrac{11}{6}\\5a+2b=\dfrac{11}{6}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a+4b=\dfrac{11}{6}\\10a+4b=\dfrac{22}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-9a=-\dfrac{11}{6}\\a+4b=\dfrac{11}{6}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{11}{54}\\b=\dfrac{\dfrac{11}{6}-\dfrac{11}{54}}{4}=\dfrac{11}{27}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\left|x+3\right|}=a=\dfrac{11}{54}\\\dfrac{1}{\left|y\right|-2}=b=\dfrac{11}{27}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}11\left|x+3\right|=54\\11\left(\left|y\right|-2\right)=27\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left|x+3\right|=\dfrac{54}{11}\\\left|y\right|=\dfrac{27}{11}+2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x+3=\dfrac{54}{11}\\x+3=\dfrac{-54}{11}\end{matrix}\right.\\\left[{}\begin{matrix}y=\dfrac{27}{11}+2\\y=-\left(\dfrac{27}{11}+2\right)\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{21}{11}\left(TM\right)\\x=\dfrac{-87}{11}\left(TM\right)\end{matrix}\right.\\\left[{}\begin{matrix}y=\dfrac{49}{11}\left(TM\right)\\y=-\dfrac{49}{11}\left(TM\right)\end{matrix}\right.\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left\{\left(\dfrac{21}{11};\dfrac{49}{11}\right);\left(\dfrac{-87}{11};\dfrac{49}{11}\right);\left(\dfrac{21}{11};\dfrac{-49}{11}\right);\left(\dfrac{-87}{11};\dfrac{-49}{11}\right)\right\}\)
DT
11 tháng 4 2022
1)
<=> \(x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
x= 0
x = 3
2) <=> \(x\left(x-3\right)=4\)
=> \(x=\dfrac{4}{x}+3\)
Y
11 tháng 4 2022
\(2,x^2-3x=4\)
\(\Leftrightarrow x^2-3x-4=0\)
\(\Delta=b^2-4ac=\left(-3\right)^2-4\left(-4\right)=25>0\)
\(\Rightarrow\)Pt có 2 nghiệm pb
\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3+5}{2}=4\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-3-5}{2}=-1\end{matrix}\right.\)
Vậy \(S=\left\{4;-1\right\}\)
\(3,x^4-5x^2+6=0\)
Đặt \(t=x^2\left(t\ge0\right)\)
Pt trở thành
\(t^2-5t+6=0\)
\(\Delta=b^2-4ac=\left(-5\right)^2-4.6=1>0\)
\(\Rightarrow\)Pt ó 2 nghiệm pb
\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{5+1}{2}=3\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-5-1}{2}-3\end{matrix}\right.\)
\(\Rightarrow t=x^2\Leftrightarrow t=\pm\sqrt{3}\)
Vậy \(S=\left\{\pm\sqrt{3}\right\}\)
\(\left(x+6\right)\left(\dfrac{x^2+3}{2}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+6=0\\\dfrac{x^2+3}{2}-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-6\\\dfrac{x^2+3}{2}=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-6\\x^2+3=2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-6\\x^2=-1\Leftrightarrow x\in\varnothing\end{matrix}\right.\\ \Leftrightarrow x=-6\)
Vậy: ...
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