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EC
2
21 tháng 5 2019
https://olm.vn/hoi-dap/tim-kiem?id=222064489607&id_subject=1&q=+++++++++++Gi%E1%BA%A3i+ph%C6%B0%C6%A1ng+tr%C3%ACnh:+x4%E2%88%922x2+7x%E2%88%9212=0++++++++++
21 tháng 5 2019
x4 - 2x2 + 7x - 12 = 0
( x4 - x3 + 3x2 ) + ( x3 - x2 + 3x ) - ( 4x2 - 4x + 12 ) = 0
x2 ( x2 - x + 3 ) + x . ( x2 - x + 3 ) - 4 ( x2 - x + 3 ) = 0
( x2 + x - 4 ) ( x2 - x + 3 ) = 0
\(\Rightarrow x^2+x-4=0\)
\(\Rightarrow x=\frac{-1\mp\sqrt{17}}{2}\)
AH
Akai Haruma
Giáo viên
23 tháng 9 2018
a)
ĐKXĐ: \(x> \frac{-5}{7}\)
Ta có: \(\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)
\(\Rightarrow 9x-7=\sqrt{7x+5}.\sqrt{7x+5}=7x+5\)
\(\Rightarrow 2x=12\Rightarrow x=6\) (hoàn toàn thỏa mãn)
Vậy......
b) ĐKXĐ: \(x\geq 5\)
\(\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}\sqrt{9}.\sqrt{x-5}=4\)
\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow 2\sqrt{x-5}=4\Rightarrow \sqrt{x-5}=2\Rightarrow x-5=2^2=4\Rightarrow x=9\)
(hoàn toàn thỏa mãn)
Vậy..........
AH
Akai Haruma
Giáo viên
23 tháng 9 2018
c) ĐK: \(x\in \mathbb{R}\)
Đặt \(\sqrt{6x^2-12x+7}=a(a\geq 0)\Rightarrow 6x^2-12x+7=a^2\)
\(\Rightarrow 6(x^2-2x)=a^2-7\Rightarrow x^2-2x=\frac{a^2-7}{6}\)
Khi đó:
\(2x-x^2+\sqrt{6x^2-12x+7}=0\)
\(\Leftrightarrow \frac{7-a^2}{6}+a=0\)
\(\Leftrightarrow 7-a^2+6a=0\)
\(\Leftrightarrow -a(a+1)+7(a+1)=0\Leftrightarrow (a+1)(7-a)=0\)
\(\Rightarrow \left[\begin{matrix} a=-1\\ a=7\end{matrix}\right.\) \(\Rightarrow a=7\) vì \(a\geq 0\)
\(\Rightarrow 6x^2-12x+7=a^2=49\)
\(\Rightarrow 6x^2-12x-42=0\Leftrightarrow x^2-2x-7=0\)
\(\Leftrightarrow (x-1)^2=8\Rightarrow x=1\pm 2\sqrt{2}\)
(đều thỏa mãn)
Vậy..........
14 tháng 7 2019
\(a,\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)\(ĐKXĐ:x\ge-\frac{5}{7}\)
\(\Leftrightarrow9x-7=7x+5\)
\(\Leftrightarrow9x-7x=5+7\)
\(\Leftrightarrow2x=12\)
\(\Leftrightarrow x=6\)
14 tháng 7 2019
\(b,\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3.\frac{\sqrt{x-5}}{\sqrt{9}}-\frac{1}{3}\sqrt{9\left(x-5\right)}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}\left(2+1-1\right)=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\)
\(\Leftrightarrow x=9\)
NN
1
NV
Nguyễn Việt Lâm
Giáo viên
31 tháng 5 2019
ĐKXĐ: \(x\ge\frac{1}{2}\)
\(\Leftrightarrow7x+3-4\sqrt{x\left(x+3\right)}-2\sqrt{2x-1}=0\)
\(\Leftrightarrow2x-1-2\sqrt{2x-1}+1+4x-4\sqrt{x\left(x+3\right)}+x+3=0\)
\(\Leftrightarrow\left(\sqrt{2x-1}-1\right)^2+\left(2\sqrt{x}-\sqrt{x+3}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2x-1}-1=0\\2\sqrt{x}-\sqrt{x+3}=0\end{matrix}\right.\) \(\Rightarrow x=1\)
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