Tìm y :
\(\left|y=\frac{1}{3}\right|-\sqrt{\frac{1}{16}}=\sqrt{\frac{1}{9}}\)
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NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 7 2020
ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\frac{1}{\sqrt[3]{x}}=a\\\frac{1}{\sqrt[3]{y}}=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^3+b^3=9\\\left(a+b\right)\left(a+1\right)\left(b+1\right)=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)^3-3ab\left(a+b\right)=9\\\left(a+b\right)\left(ab+a+b+1\right)=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)^3-3ab\left(a+b\right)=9\\ab\left(a+b\right)+\left(a+b\right)^2+\left(a+b\right)=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)^3-3ab\left(a+b\right)=9\\3ab\left(a+b\right)+3\left(a+b\right)^2+3\left(a+b\right)=54\end{matrix}\right.\)
\(\Rightarrow\left(a+b\right)^3+3\left(a+b\right)^2+3\left(a+b\right)=63\)
\(\Leftrightarrow\left(a+b\right)^3+3\left(a+b\right)^2+3\left(a+b\right)+1=64\)
\(\Leftrightarrow\left(a+b+1\right)^3=4^3\)
\(\Leftrightarrow a+b+1=4\Rightarrow a+b=3\)
\(\Rightarrow3\left(ab+3+1\right)=18\Rightarrow ab=2\)
Theo Viet đảo; a và b là nghiệm:
\(t^2-3t+2=0\Rightarrow\left[{}\begin{matrix}t=1\\t=2\end{matrix}\right.\)
\(\Rightarrow\left(a;b\right)=\left(1;2\right);\left(2;1\right)\Rightarrow\left(x;y\right)=\left(1;\frac{1}{8}\right);\left(\frac{1}{8};1\right)\)
30 tháng 7 2019
chịu thua vô điều kiện xin lỗi nha : v
30 tháng 7 2019
muốn biết câu trả lời lo mà sệt trên google ấy đừng có mà dis:v
21 tháng 10 2016
Bài 1
a, \(\left(\frac{\sqrt{y}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\frac{\sqrt{x}\left(\sqrt{y}-1\right)}{\sqrt{y}-1}\right).\sqrt{y}\left(\sqrt{x}-1\right)\)
=\(\left(\sqrt{y}+\sqrt{x}\right).\sqrt{y}\left(\sqrt{x}-1\right)\)
b,\(\sqrt{8+2.2\sqrt{2}+1}-\sqrt{8-2.2\sqrt{2}+1}\)
=\(\sqrt{\left(\sqrt{8}+1\right)^2}-\sqrt{\left(\sqrt{8}-1\right)^2}\)
=\(\sqrt{8}+1-\left(\sqrt{8}-1\right)\)
=2
Bài 2
a, ĐKXĐ : x\(\ge\)0, x\(\pm\)1
b, Q=\(\left(\frac{\sqrt{x}\left(1+\sqrt{x}\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}+\frac{\sqrt{x}\left(1-\sqrt{x}\right)}{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}\right)+\frac{3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\left(\frac{\sqrt{x}\left(1+\sqrt{x}\right)+\sqrt{x}\left(1-\sqrt{x}\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\right)+\frac{3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\left(\frac{\sqrt{x}+x+\sqrt{x}-x}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\right)+\frac{3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{2\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}-\frac{3-\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)
=\(\frac{2\sqrt{x}-3+\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)
=\(\frac{3\sqrt{x}-3}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)
=\(\frac{-3}{1+\sqrt{x}}\)
c, de Q = 2 => \(\frac{-3}{1+\sqrt{x}}\)=2 =>1+\(\sqrt{x}\)=-6 =>\(\sqrt{x}\)=-7 =>x vô nghiệm
21 tháng 10 2016
Bài 1: \(\left(\frac{\sqrt{xy}-\sqrt{y}}{\sqrt{x}-1}+\frac{\sqrt{xy}-\sqrt{x}}{\sqrt{y}-1}\right)\cdot\left(\sqrt{xy}-\sqrt{y}\right)\)
\(=\left(\frac{\sqrt{y}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\frac{\sqrt{x}\left(\sqrt{y}-1\right)}{\sqrt{y}-1}\right)\cdot\left(\sqrt{xy}-\sqrt{y}\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{xy}-\sqrt{y}\right)\)
\(\sqrt{9+4\sqrt{2}}-\sqrt{9-4\sqrt{2}}=\sqrt{\left(2\sqrt{2}+1\right)^2}-\sqrt{\left(2\sqrt{2}-1\right)^2}\\ =2\sqrt{2}+1-2\sqrt{2}+1=2\)
21 tháng 10 2016
Bài 2:
\(Q=\left(\frac{\sqrt{x}}{1-\sqrt{x}}+\frac{\sqrt{x}}{1+\sqrt{x}}\right)+\frac{3-\sqrt{x}}{x-1}\left(ĐK:x\ge0;x\ne1\right)\)
\(=\frac{-\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}\left(\sqrt{x}-1\right)+3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-x-\sqrt{x}+x-\sqrt{x}+3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{-3}{\sqrt{x}+1}\)
Để Q=2
=> \(\frac{-3}{\sqrt{x}+1}=2\)
\(\Leftrightarrow2\left(\sqrt{x}+1\right)=-3\)
\(\Leftrightarrow2\sqrt{x}+2=-3\)
\(\Leftrightarrow2\sqrt{x}=-5\) (vô lí)
Vậy k có giá trị nào của x thỏa mãn Q=2
Dấu = là sau y là sai rồi bạn
Là dấu trừ "-"