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AH
Akai Haruma
Giáo viên
5 tháng 6 2018
Bài 1:
\(\frac{(x+1)^4}{(x^2+1)^2}+\frac{4x}{x^2+1}=6\)
\(\Leftrightarrow \frac{(x+1)^4+4x(x^2+1)}{(x^2+1)^2}=6\)
\(\Leftrightarrow \frac{x^4+8x^3+6x^2+8x+1}{(x^2+1)^2}=6\Rightarrow x^4+8x^3+6x^2+8x+1=6(x^2+1)^2\)
\(\Leftrightarrow x^4+8x^3+6x^2+8x+1=6(x^4+2x^2+1)\)
\(\Leftrightarrow 5x^4-8x^3+6x^2-8x+5=0\)
\(\Leftrightarrow 5x^3(x-1)-3x^2(x-1)+3x(x-1)-5(x-1)=0\)
\(\Leftrightarrow (x-1)(5x^3-3x^2+3x-5)=0\)
\(\Leftrightarrow (x-1)[5(x-1)(x^2+x+1)-3x(x-1)]=0\)
\(\Leftrightarrow (x-1)^2(5x^2+2x+5)=0\)
Dễ thấy \(5x^2+2x+5>0\), do đó \((x-1)^2=0\Leftrightarrow x=1\)
AH
Akai Haruma
Giáo viên
5 tháng 6 2018
Bài 2: ĐK: \(x\geq 0\)
\(A=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1\)
\(A=\frac{\sqrt{x}(\sqrt{x^3}-1)}{x+\sqrt{x}+1}-\frac{\sqrt{x}(\sqrt{x^3}+1)}{x-\sqrt{x}+1}+x+1\)
\(A=\frac{\sqrt{x}(\sqrt{x}-1)(x+\sqrt{x}+1)}{x+\sqrt{x}+1}-\frac{\sqrt{x}(\sqrt{x}+1)(x-\sqrt{x}+1)}{x-\sqrt{x}+1}+x+1\)
\(A=\sqrt{x}(\sqrt{x}-1)-\sqrt{x}(\sqrt{x}+1)+x+1\)
\(A=x-2\sqrt{x}+1=(\sqrt{x}-1)^2\)
PA
18 tháng 8 2017
\(\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}=m\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(2-\sqrt{x-4}\right)^2}=m\)
\(\Leftrightarrow\left|\sqrt{x-4}+2\right|+\left|2-\sqrt{x-4}\right|=m\)
mà \(\left|\sqrt{x-4}+2\right|+\left|2-\sqrt{x-4}\right|\)
\(\ge\left|\sqrt{x-4}+2+2-\sqrt{x-4}\right|=4\)
\(\Rightarrow m\ge4\) thì pt trên có no
NT
0
PT
21 tháng 3 2017
ta thấy pt luôn có no . Theo hệ thức Vi - ét ta có:
x1 + x2 = \(\dfrac{-b}{a}\) = 6
x1x2 = \(\dfrac{c}{a}\) = 1
a) Đặt A = x1\(\sqrt{x_1}\) + x2\(\sqrt{x_2}\) = \(\sqrt{x_1x_2}\)( \(\sqrt{x_1}\) + \(\sqrt{x_2}\) )
=> A2 = x1x2(x1 + 2\(\sqrt{x_1x_2}\) + x2)
=> A2 = 1(6 + 2) = 8
=> A = 2\(\sqrt{3}\)
b) bạn sai đề
19 tháng 7 2017
a. \(ĐKXĐ\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
b. \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)-2-5\sqrt{x}}{x-4}\)
\(=\dfrac{3x-6\sqrt{x}}{x-4}=\dfrac{3\sqrt{x}\left(\sqrt{x-2}\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
c. \(P=2\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}+2}=2\Leftrightarrow3\sqrt{x}=2\sqrt{x}+4\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\)
Vậy x = 16
14 tháng 10 2018
Bo may la binh day k di hieu ashdbfgbgygygggydfsghuyfhdguuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu3
11 tháng 7 2017
(1) \(\Leftrightarrow\left(x+1\right)\left(\sqrt{16x+17}-x+\dfrac{23}{8}\right)=0\)
cái này đâu ra z ???
CW
11 tháng 7 2017
nguyen van tuan: hì, xin lỗi, làm hơi tắt ^^!
\(\left(1\right)\Leftrightarrow\left(x+1\right)\sqrt{16x+17}=\left(x+1\right)\left(x-\dfrac{23}{8}\right)\Leftrightarrow\left(x+1\right)\sqrt{16x+17}-\left(x+1\right)\left(x-\dfrac{23}{8}\right)=0\Leftrightarrow\left(x+1\right)\left(\sqrt{16x+17}-x+\dfrac{23}{8}\right)=0\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\x-\dfrac{1}{x}\ge0\end{matrix}\right.\)
Pt\(\Rightarrow x^2-2x+1=2-x\sqrt{x-\dfrac{1}{x}}\)
\(\Rightarrow x^2-2x-1+x\sqrt{x-\dfrac{1}{x}}=0\)
\(\Rightarrow x-2-\dfrac{1}{x}+\sqrt{x-\dfrac{1}{x}}=0\)
Đặt \(\sqrt{x-\dfrac{1}{x}}=a\ge0\)
\(\Rightarrow a^2+a-2=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-2\left(ktm\right)\end{matrix}\right.\)
\(\Rightarrow x-\dfrac{1}{x}=1\) (thỏa ĐKXĐ)
\(\Rightarrow x^2-x-1=0\)
\(\Rightarrow x=\dfrac{1\pm\sqrt{5}}{2}\)