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NM
0
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
\(\Rightarrow x^2+x-4=0\)
\(\Rightarrow x=\frac{-1\mp\sqrt{17}}{2}\)
NN
1
NV
Nguyễn Việt Lâm
Giáo viên
31 tháng 5 2019
ĐKXĐ: \(x\ge\frac{1}{2}\)
\(\Leftrightarrow7x+3-4\sqrt{x\left(x+3\right)}-2\sqrt{2x-1}=0\)
\(\Leftrightarrow2x-1-2\sqrt{2x-1}+1+4x-4\sqrt{x\left(x+3\right)}+x+3=0\)
\(\Leftrightarrow\left(\sqrt{2x-1}-1\right)^2+\left(2\sqrt{x}-\sqrt{x+3}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2x-1}-1=0\\2\sqrt{x}-\sqrt{x+3}=0\end{matrix}\right.\) \(\Rightarrow x=1\)
KN
2
CT
16 tháng 6 2019
Trả lời
đưa căn 7x+2 sang vế bên phải rồi mũ 3 lên là đc mầ
hok tốt
16 tháng 6 2019
ĐKXĐ: x > -2/7
Đặt \(\hept{\begin{cases}\sqrt[3]{2x-1}=a\\\sqrt{7x+2}=b\ge0\end{cases}}\Rightarrow7a^3-2b^2=14x-7-14x-4=-11\)
Từ đề bài \(\Rightarrow4a-b=1\)
Ta có hệ \(\hept{\begin{cases}7a^3-2b^2=-11\\4a-b=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}7a^3-2b^2=-11\\b=4a-1\end{cases}}\)
\(\Rightarrow7a^3-2\left(4a-1\right)^2=-11\)
\(\Leftrightarrow\left(a-1\right)\left(7a^2-25a-9\right)=0\)
Đến đây tìm được a => x
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gì đây?