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a) \(\frac{4x}{\sqrt{7x-6}}+\frac{4\sqrt{7x-6}}{x}=8\) Đặt \(\frac{x}{\sqrt{7x-6}}=t\left(ĐK:t\ge0\right)\Leftrightarrow\frac{1}{t}=\frac{\sqrt{7x-6}}{x}\\ Pt\Leftrightarrow4t+\frac{4}{t}=8\Leftrightarrow4t^2+4-8t=0\Leftrightarrow t=1\left(tm\right)\)
Với
\(t=1\Leftrightarrow\frac{x}{\sqrt{7x-6}}=1\Leftrightarrow x=\sqrt{7x-6}\Leftrightarrow x^2=7x-6\Leftrightarrow x^2-7x+6=0\Leftrightarrow\left[\begin{array}{nghiempt}x=6\\x=1\end{array}\right.\)
Vậy \(s=\left\{1;6\right\}\)
27>25>0
→\(\sqrt{27}\)>\(\sqrt{25}\)
\(\sqrt{27}\)>5
6>4>0
\(\sqrt{6}\)>\(\sqrt{4}\)
\(\sqrt{6}\)>2
\(\sqrt{27}\)+\(\sqrt{6}\)>2+5→\(\sqrt{27}\)+\(\sqrt{6}\)>7
0<48<49→\(\sqrt{48}\)<\(\sqrt{49}\)→\(\sqrt{48}\)<7
Từ đó suy ra \(\sqrt{27}\)+\(\sqrt{6}\)>\(\sqrt{48}\)
a) Đặt \(t=\sqrt{2x^2-3x+5}\ge0\) thì
\(2t=t^2-11\)
\(\Leftrightarrow\left[{}\begin{matrix}t=1+2\sqrt{3}\\t=1-2\sqrt{3}\end{matrix}\right.\)
Vì \(t\ge0\) nên \(t=1+2\sqrt{3}\)
\(\Rightarrow\sqrt{2x^2-3x+5}=1+2\sqrt{3}\)
\(\Leftrightarrow2x^2-3x+5=13-4\sqrt{3}\)
\(\Leftrightarrow2x^2-3x-8+4\sqrt{3}=0\)
Giải pt trên tìm được x
c) ĐK: \(x\ge0\)
Đặt \(a=\sqrt{x}\ge0;b=\sqrt{x+3}\ge0\)
pt trên đc viết lại thành
\(2b^2+2ab=4\left(a+b\right)\)
\(\Leftrightarrow\left(b-2\right)\left(a+b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}b=2\\a=-b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=2\\\sqrt{x}=-\sqrt{x+3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=x+3\end{matrix}\right.\)
Vậy pt có 1 nghiệm duy nhất x = 1.
b) ĐK: tự làm
Ta có \(\left(x+5\right)\left(2-x\right)=-x\left(x+3\right)+10\)
Đặt \(a=\sqrt{x}\ge0;b=\sqrt{x+3}\ge0\)
pt trên đc viết lại thành
\(-a^2b^2+10=3ab\)
\(\Leftrightarrow-a^2b^2-3ab+10=0\) (*)
Đặt \(t=ab\ge0\) thì (*) \(\Rightarrow-t^2-3t+10=0\)
\(\Leftrightarrow\left[{}\begin{matrix}ab=t=2\\ab=t=-5\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{x\left(x+3\right)}=2\)
Bạn tự làm tiếp nhé
\(A=\left(\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\left(ĐK:x>0;x\ne1;x\ne4\right)\)
\(=\frac{\sqrt{x}-\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}:\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{2}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{x-1-x+4}\)
\(=\frac{2\left(\sqrt{x}+1\right)}{3\sqrt{x}}\)
Điều kiện: \(\left\{\begin{matrix}x\ge0\\y\ge1\\z\ge2\end{matrix}\right.\)
Ta có: \(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{x+y+z}{2}\)
\(\Leftrightarrow-2\sqrt{x}-2\sqrt{y-1}-2\sqrt{z-2}+x+y+z=0\)
\(\Leftrightarrow\left(x-2\sqrt{x}+1\right)+\left(y-1-2\sqrt{y-1}+1\right)+\left(z-2-2\sqrt{z-2}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-1\right)^2=0\)
\(\Leftrightarrow\left\{\begin{matrix}\sqrt{x}=1\\\sqrt{y-1}=1\\\sqrt{z-2}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{\begin{matrix}x=1\\y=2\\z=3\end{matrix}\right.\)
\(\Rightarrow x_0^2+y_0^2+z_0^2=1^2+2^2+3^2=14\)
a: \(D=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
b: \(E=\sqrt{6-2\sqrt{5}}\cdot\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=18+6\sqrt{5}-6\sqrt{5}-10=8\)
1a) \(\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(=\sqrt{4+\sqrt{8}}.\sqrt{\left(2+\sqrt{2+\sqrt{2}}\right)\left(\sqrt{2-\sqrt{2+\sqrt{2}}}\right)}\)
\(=\sqrt{4+\sqrt{8}}.\sqrt{4-2-\sqrt{2}}\)
\(=\sqrt{4+\sqrt{8}}.\sqrt{2-\sqrt{2}}=\sqrt{\left(4+\sqrt{8}\right)\left(2-\sqrt{2}\right)}\)
\(=\sqrt{8-4\sqrt{2}-\sqrt{16}+2\sqrt{8}}\)
\(=\sqrt{8-4\sqrt{2}-4+4\sqrt{2}}\)
\(=\sqrt{4}=2\)
1b) \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+4\sqrt{3}+3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{25-10\sqrt{3}+3}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)
\(=\sqrt{25}=5\)
a/ \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\) \(=\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}=-2\sqrt{3}\).
b/ \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\Rightarrow A^2=8+2\sqrt{4^2-\left(\sqrt{10+2\sqrt{5}}\right)^2}=8+2\sqrt{6-2\sqrt{5}}\) \(=8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\sqrt{5}-2=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\)
\(\Rightarrow A=\sqrt{5}+1\)
c/ \(B=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\Rightarrow\sqrt{2}B=\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-2\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\sqrt{5}+2=2\Rightarrow B=\sqrt{2}\)
binh phuong len
\(\left(\sqrt{8}+\sqrt{5}\right)^2=13+2\sqrt{40}\)
\(\left(\sqrt{7}+\sqrt{6}\right)^2=13+2\sqrt{42}\)
vi
\(2\sqrt{40}< 2\sqrt{42}\)
nen \(\sqrt{8}+\sqrt{5}< \sqrt{7}+\sqrt{6}\)