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Bài 3: \(3\left(\sqrt{2x^2+1}-1\right)=x\left(1+3x+8\sqrt{2x^2+1}\right)\)
\(\Leftrightarrow\left(3-8x\right)\sqrt{2x^2+1}=3x^2+x+3\)
\(\Rightarrow\left(3-8x\right)^2\left(2x^2+1\right)=\left(3x^2+x+3\right)^2\)
\(\Leftrightarrow119x^4-102x^3+63x^2-54x=0\)
\(\Leftrightarrow x\left(7x-6\right)\left(17x^2+9\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{6}{7}\end{cases}}\)
Thử lại, ta nhận được \(x=0\)là nghiệm duy nhất của phương trình
ai nay dung kinh nghiem la chinh
cau a)
ta thay \(10+6\sqrt{3}=\left(1+\sqrt{3}\right)^3\)
\(6+2\sqrt{5}=\left(1+\sqrt{5}\right)^2\)
khi do \(x=\frac{\sqrt[3]{\left(\sqrt{3}+1\right)^3}\left(\sqrt{3}-1\right)}{\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{5}}\)
\(x=\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{1+\sqrt{5}-\sqrt{5}}\)
\(x=\frac{3-1}{1}=2\)
suy ra
x^3-4x+1=1
A=1^2018
A=1
b)
ta thay
\(7+5\sqrt{2}=\left(1+\sqrt{2}\right)^3\)
khi do
\(x=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\frac{1}{\sqrt[3]{\left(1+\sqrt{2}\right)^3}}\)
\(x=1+\sqrt{2}-\frac{1}{1+\sqrt{2}}=\frac{\left(1+\sqrt{2}\right)^2-1}{1+\sqrt{2}}=\frac{2+2\sqrt{2}}{1+\sqrt{2}}\)
x=2
thay vao
x^3+3x-14=0
B=0^2018
B=0
a) Rút gọn : Q =\(\left(\frac{\sqrt{x}-3}{\sqrt{x}+3}+\frac{\sqrt{x}+3}{\sqrt{x}-3}-\frac{14}{9-x}\right).\frac{\sqrt{x}-3}{2}\left(x\ge0,x\ne9\right)\)
Q =\(\left(\frac{\sqrt{x}-3}{\sqrt{x}+3}+\frac{\sqrt{x}+3}{\sqrt{x}-3}+\frac{14}{x-9}\right).\frac{\sqrt{x}-3}{2}\)
Q =\(\left(\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{14}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right).\frac{\sqrt{x}-3}{2}\)
Q = \(\frac{x-6\sqrt{x}+9+x+6\sqrt{x}+9+14}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{2}\)
Q = \(\frac{2x+32}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{2}\)
Q = \(\frac{2\left(x+16\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{2}\)
Q = \(\frac{x+16}{\sqrt{x}+3}\)
thay \(x=7-4\sqrt{3}\) vào Q ta được
Q =\(\frac{7-4\sqrt{3}+16}{\sqrt{7-4\sqrt{3}}+3}\) =\(\frac{23-4\sqrt{3}}{\sqrt{\left(2-\sqrt{3}\right)^2+3}}\)
=\(\frac{23-4\sqrt{3}}{2-\sqrt{3}+3}\)
=\(\frac{23-4\sqrt{3}}{5-\sqrt{3}}\)
a, Q = \(\left(\frac{\sqrt{x}-3}{\sqrt{x}+3}+\frac{\sqrt{x}+3}{\sqrt{x}-3}-\frac{14}{9-x}\right)\times\frac{\sqrt{x}-3}{2}\)
= \(\left[\frac{\left(\sqrt{x}-3\right)^2+\left(\sqrt{x}+3\right)^2+14}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]\times\frac{\sqrt{x}-3}{2}\)
= \(\left[\frac{x-6\sqrt{x}+9+x+6\sqrt{x}+9+14}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]\times\frac{\sqrt{x}-3}{2}\)
= \(\frac{2x+32}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\times\frac{\sqrt{x}-3}{2}\)
= \(\frac{2\left(x+16\right)\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
= \(\frac{x+16}{\sqrt{x}+3}\)
Thay \(x=7-4\sqrt{3}\) vào Q ta được:
Q= \(\frac{7-4\sqrt{3}+16}{\sqrt{7-4\sqrt{3}}+3}\) = \(\frac{23-4\sqrt{3}}{\sqrt{\left(2-\sqrt{3}\right)^2}+3}\)=\(\frac{23-4\sqrt{3}}{2+3-\sqrt{3}}=\frac{23-4\sqrt{3}}{5-\sqrt{3}}=\frac{\left(23-4\sqrt{3}\right)\left(5+\sqrt{3}\right)}{\left(5+\sqrt{3}\right)\left(5-\sqrt{3}\right)}\) =\(\frac{103+3\sqrt{3}}{22}\)
b,
\(Q=\frac{x+16}{\sqrt{x}+3}=\frac{x+9+7}{\sqrt{x}+3}=2+\frac{7}{\sqrt{x}+3}\)
Ta có \(2+\frac{7}{\sqrt{x}+3}\) nhỏ nhất khi \(\sqrt{x}+3\) nhỏ nhất
Mà với điều kiện \(x\ge0\) nên GTNNQ=\(2+\frac{7}{3}=\frac{13}{3}\)
chịu thua vô điều kiện xin lỗi nha : v
muốn biết câu trả lời lo mà sệt trên google ấy đừng có mà dis:v
\(P=\frac{1}{\sqrt{x}+1}+\frac{10}{2\sqrt{x}+1}-\frac{5}{2x+3\sqrt{x}+1}\)
\(=\frac{1}{\sqrt{x}+1}+\frac{10}{2\sqrt{x}+1}-\frac{5}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2\sqrt{x}+1+10\left(\sqrt{x}+1\right)-5}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2\sqrt{x}+1+10\sqrt{x}+10-5}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{6}{\sqrt{x}+1}\)
b) Để P nguyên tố thì \(\frac{6}{\sqrt{x}+1}\) nguyên tố
Để \(P\inℕ^∗\) thì \(\sqrt{x}+1\inƯ\left(6\right)\)
Mà P nguyên tố \(\Rightarrow\frac{6}{\sqrt{x}+1}=\left\{2;3\right\}\Rightarrow\sqrt{x}+1=\left\{2;3\right\}\)
Với \(\sqrt{x}+1=2\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)
Với \(\sqrt{x}+1=3\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)
Vậy ...........
\(x=\dfrac{1}{\sqrt{2}}\left(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\right)=\sqrt{6}\)
\(y=\sqrt{\left(\sqrt{6}-1\right)^2}=\sqrt{6}-1\)
\(\Rightarrow x-y=1\Rightarrow P=1\)
\(B=x-2020-\sqrt{x-2020}+\dfrac{1}{4}+\dfrac{8079}{4}\)
\(B=\left(\sqrt{x-2020}-\dfrac{1}{2}\right)^2+\dfrac{8079}{4}\ge\dfrac{8079}{4}\)
\(B_{min}=\dfrac{8079}{4}\) khi \(x=\dfrac{8081}{4}\)