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a. \(\sqrt{-2x+3}\)
ĐKXĐ: x < 0
b. \(\sqrt{\dfrac{2}{x^2}}\)
ĐKXĐ: x \(\ne\) 0
c. \(\sqrt{\dfrac{4}{x+3}}\)
ĐKXĐ: x > -3
d. \(\sqrt{\dfrac{-5}{x^2+6}}\)
ĐKXĐ: x vô nghiệm
4. a. x2 - 7
= x2 - \(\left(\sqrt{7}\right)^2\)
= \(\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)
b. x2 - \(2\sqrt{2}x\) + 2
= x2 - \(2\sqrt{2}x\) + \(\left(\sqrt{2}\right)^2\)
= (x - \(\sqrt{2}\))2
c. x2 + \(2\sqrt{13}x\) + 13
= x2 + \(2\sqrt{13}x\) + \(\left(\sqrt{13}\right)^2\)
= \(\left(x+\sqrt{13}\right)^2\)
sin 650=cos 350
\(cos70^0=sin30^0\)
\(tan80^0=cot20^0\)
\(cot68^0=tan32^0\)
x2 - (m-1)x + 2m-6 = 0
a)xét delta
(m-1)2 - 4(2m-6) = m2 - 2m + 1 - 8m + 24
= m2 - 10m + 25 = (m-5)2 ≥ 0
=> pt luôn có 2 nghiệm với mọi m thuộc R
b) theo Vi-ét ta có
\(\left\{{}\begin{matrix}x1+x2=m-1\\x1x2=2m-6\end{matrix}\right.\)
theo đề ta có \(A=\dfrac{2x1}{x2}+\dfrac{2x2}{x1}\) đk: m ≠ 3
A = \(\dfrac{2x1^2+2x2^2}{x1x2}=\dfrac{2\left(\left(x1+x2\right)^2-2x1x2\right)}{2m-6}\)
A=\(\dfrac{m^2-6m+25}{m-3}\)
để A có giá trị nguyên thì m2 - 6m + 25 ⋮ m - 3
m2 - 6m + 9 + 16 ⋮ m - 3
(m-3)2 + 16 ⋮ m-3
16 ⋮ m - 3 => m-3 thuộc ước của 16
U(16) = { - 16; - 8; - 4; -2 ; -1 ; 1 ; 2; 4; 8; 16 }
=> m- 3 = { - 16; - 8; - 4; -2 ; -1 ; 1 ; 2; 4; 8; 16 }
m = { - 13 ; -5 ; -1; 1; 2; 4; 5; 7; 11; 19 }
\(\Rightarrow\left\{{}\begin{matrix}x+my=m+1\\m^2x+my=2m^2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+my=m+1\\\left(m^2-1\right)x=2m^2-m-1\end{matrix}\right.\)
Phương trình có nghiệm duy nhất khi \(m^2-1\ne0\Rightarrow m\ne\pm1\)
Khi đó ta có: \(x=\dfrac{2m^2-m-1}{m^2-1}=\dfrac{\left(m-1\right)\left(2m+1\right)}{\left(m-1\right)\left(m+1\right)}=\dfrac{2m+1}{m+1}\)
\(\Rightarrow y=2m-mx=\dfrac{m}{m+1}\)
Để \(\left\{{}\begin{matrix}x\ge2\\y\ge1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{2m+1}{m+1}\ge2\\\dfrac{m}{m+1}\ge1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{-1}{m+1}\ge0\\\dfrac{-1}{m+1}\ge0\end{matrix}\right.\)
\(\Rightarrow m+1< 0\Rightarrow m< -1\)
a, \(\dfrac{\sqrt{80}}{\sqrt{5}}\)-\(\sqrt{5}\).\(\sqrt{20}\)= \(\sqrt{16}\)-10=-6
b, (\(\sqrt{28}\)-\(\sqrt{12}\)-\(\sqrt{7}\))\(\sqrt{7}\)+2\(\sqrt{21}\)=\(\sqrt{196}\)-\(\sqrt{84}\)-7+2 \(\sqrt{21}\)=14-7=7
c, \(\sqrt[3]{2}\).\(\sqrt[3]{32}\)+\(\sqrt{2}\).\(\sqrt{32}\)=\(\sqrt[3]{64}\)+\(\sqrt{64}\)=4+8=12
d, \(2\sqrt{8\sqrt{3}}\)-\(\sqrt{2\sqrt{3}}\)-\(\sqrt{9\sqrt{12}}\)=\(4\sqrt{12}\)-\(\sqrt{12}\)-\(3\sqrt{12}\)=0
1: Thay x=16 vào A, ta được:
\(A=\dfrac{16-4}{4-2}=\dfrac{12}{2}=6\)
2: \(B=\dfrac{x-4+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}=\dfrac{x-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)
a: \(x^2-7=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)
b: \(x^2-2x\sqrt{2}+2=\left(x-\sqrt{2}\right)^2\)
c: \(x^2+2x\sqrt{13}+13=\left(x+\sqrt{13}\right)^2\)