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Câu 1: Đề bài sai, với điều kiện đề bài đã cho thì Q vẫn nguyên tại \(x=0\), đề bài đúng phải là \(\forall x>0\) thì Q không nguyên (ko hiểu sao lại có điều kiện \(x\ne4\) , cái này hoàn toàn ko ảnh hưởng gì tới bài toán)
\(A=Q^2=\frac{x+4\sqrt{x}+4}{x+4}\Leftrightarrow Ax+4A=x+4\sqrt{x}+4\)
\(\Leftrightarrow\left(A-1\right)x-4\sqrt{x}+4A-4=0\)
\(\Delta'=4-\left(4A-4\right)\left(A-1\right)\ge0\)
\(\Leftrightarrow=-A^2+2A\ge0\Rightarrow0\le A\le2\Rightarrow A\le2\)
\(\Rightarrow Q\le\sqrt{2}< 2\)
Mặt khác ta có \(\sqrt{x}+2=\sqrt{x}+\sqrt{4}>\sqrt{x+4}\)
\(\Rightarrow Q=\frac{\sqrt{x}+2}{\sqrt{x+4}}>1\) \(\Rightarrow1< Q< 2\Rightarrow Q\) không thể nhận giá trị nguyên
Câu 2: ĐKXĐ: \(x\ge-2\)
a/ \(\Leftrightarrow4\left(x^2+2x+3\right)+3\left(x+2\right)=8\sqrt{\left(x+2\right)\left(x^2+2x+3\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+2}=a\ge0\\\sqrt{x^2+2x+3}=b>0\end{matrix}\right.\) ta được:
\(3a^2-8ab+4b^2=0\Leftrightarrow\left(a-2b\right)\left(3a-2b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2b\\3a=2b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+2}=2\sqrt{x^2+2x+3}\\3\sqrt{x+2}=2\sqrt{x^2+2x+3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x^2+7x+10=0\left(vn\right)\\4x^2-x-6=0\end{matrix}\right.\) \(\Rightarrow x=\frac{1\pm\sqrt{97}}{8}\)
b/ ĐKXĐ: \(\left[{}\begin{matrix}x\ge7\\-5\le x\le-2\end{matrix}\right.\)
\(\Leftrightarrow3x^2-11x-22=7\sqrt{\left(x^2-5x-14\right)\left(x+5\right)}\)
\(\Leftrightarrow3\left(x^2-5x-14\right)+4\left(x+5\right)-7\sqrt{\left(x^2-5x-14\right)\left(x+5\right)}=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-5x-14}=a\ge0\\\sqrt{x+5}=b\ge0\end{matrix}\right.\) ta được:
\(3a^2-7ab+4b^2=0\Leftrightarrow\left(a-b\right)\left(3a-4b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\3a=4b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-5x-14}=\sqrt{x+5}\\3\sqrt{x^2-5x-14}=4\sqrt{x+5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-6x-19=0\\9x^2-61x-206=0\end{matrix}\right.\) \(\Rightarrow x=...\)
1.
\(\sqrt{\dfrac{x-1+\sqrt{2x-3}}{x+2-\sqrt{2x+3}}}\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\sqrt{\dfrac{\left(\sqrt{2x-3}+1\right)^2}{\left(\sqrt{2x+3}-1\right)^2}}\end{matrix}\right.\)\(\Leftrightarrow\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{2x-3}+1}{\sqrt{2x+3}-1}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\left(\sqrt{2x-3}+1\right)\left(\sqrt{2x+3}+1\right)}{2\left(x+1\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{4x^2-9}+\sqrt{2x-3}+\sqrt{2x+3}+1}{2\left(x+1\right)}\end{matrix}\right.\)
hết tối giải rồi
ráng làm nốt rồi đi ngủ thoyy
1.
a) ĐK: \(x\ge2\)
\(\sqrt{x^2-3x+2}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{x^2+2x-3}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{\left(x+3\right)\left(x-1\right)}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x+3}-\sqrt{x-2}-\sqrt{\left(x+3\right)\left(x-1\right)}\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x-2}-\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x-2}=\sqrt{x+3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-2=x+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\varnothing\end{matrix}\right.\)
Vậy...
b) \(\left(4x+2\right)\sqrt{x+8}=3x^2+7x+8\)
\(\Leftrightarrow2\left(2x+1\right)\sqrt{x+8}=4x^2+4x+1+x+8-x^2+2x-1\)
\(\Leftrightarrow2\left(2x+1\right)\sqrt{x+8}=\left(2x+1\right)^2+\left(x+8\right)-\left(x-1\right)^2\)
\(\Leftrightarrow\left(2x+1\right)^2-2\left(2x-1\right)\sqrt{x+8}+\left(x+8\right)-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(2x+1-\sqrt{x+8}\right)^2-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(2x+1-\sqrt{x+8}-x+1\right)\left(2x+1-\sqrt{x+8}+x-1\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{x+8}+2\right)\left(3x-\sqrt{x+8}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=\sqrt{x+8}\\3x=\sqrt{x+8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=1\end{matrix}\right.\)\(\Leftrightarrow x=1\)
Vậy...
c) \(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)
Nhân cả 2 vế với \(\sqrt{2}\) ta được :
\(pt\Leftrightarrow\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x-2\sqrt{2x-1}}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x-1}-1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}-1\right|=2\)
Ta có : \(\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}-1\right|\)
\(=\left|\sqrt{2x-1}+1\right|+\left|1-\sqrt{2x-1}\right|\ge\left|\sqrt{2x-1}+1+1-\sqrt{2x-1}\right|=2\)
Dấu "=" xảy ra \(\Leftrightarrow\left(\sqrt{2x-1}+1\right)\left(1-\sqrt{2x-1}\right)\ge0\Leftrightarrow\frac{1}{2}\le x\le1\)
2) \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right):\frac{1}{x+y+z}=1\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{x+y+z}-\frac{1}{z}\)
\(\Leftrightarrow\frac{x+y}{xy}=\frac{z-x-y-z}{z\left(x+y+z\right)}\)
\(\Leftrightarrow\frac{x+y}{xy}=\frac{-\left(x+y\right)}{z\left(x+y+z\right)}\)
\(\Leftrightarrow z\left(x+y\right)\left(x+y+z\right)=-xy\cdot\left(x+y\right)\)
\(\Leftrightarrow\left(x+y\right)\left(xz+yz+z^2+xy\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\z+x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)
TH1: \(x=-y\Leftrightarrow x^{29}=-y^{29}\Leftrightarrow x^{29}+y^{29}=0\)
Khi đó \(B=0\cdot\left(x^{11}+y^{11}\right)\cdot\left(x^{2013}+y^{2013}\right)=0\)
Tương tự 2 trường hợp còn lại ta đều được \(B=0\)
Vậy \(B=0\)
a,Ta có :\(x=\sqrt[3]{4\left(\sqrt{5}+1\right)}-\sqrt[3]{4\left(\sqrt{5}-1\right)}\)
\(\Rightarrow x^3=4\left(\sqrt{5}+1\right)-4\left(\sqrt{5}-1\right)-3\sqrt[3]{4\left(\sqrt{5}-1\right).4\left(\sqrt{5}+1\right)}.\left(\sqrt[3]{4\left(\sqrt{5}+1\right)}-\sqrt[3]{4\left(\sqrt{5}-1\right)}\right)\)\(\Rightarrow x^3=8-3\sqrt[3]{16\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}.x\)
\(\Rightarrow x^3=8-3\sqrt[3]{64}.x\Rightarrow x^3=8-12x\)\(\Rightarrow x^3-12x+8=0\)
Vậy \(x^3+12x-8=0\)
b,\(\left(x+\sqrt{x^2+3}\right)\left(y+\sqrt{y^2+3}\right)=3\)(1)
Ta có :\(3=\left(x^2+3\right)-x^2=\left(\sqrt{x^2+3}-x\right)\left(\sqrt{x^2+3}+x\right)\)(2)
\(3=\left(y^2+3\right)-y^2=\left(\sqrt{y^2+3}-y\right)\left(\sqrt{y^2+3}+y\right)\) (3)
Từ (1) và (2) ta suy ra :\(y+\sqrt{y^2+3}=\sqrt{x^2+3}-x\)
Từ (1) và (3) ta suy ra :\(x+\sqrt{x^2+3}=\sqrt{y^2+3}-y\)
Cộng 2 đẳng thức trên vế theo vế ta được :
\(x+y+\sqrt{x^2+3}+\sqrt{y^2+3}=\sqrt{x^2+3}+\sqrt{y^2+3}-x-y\)
\(\Leftrightarrow2\left(x+y\right)=0\Leftrightarrow x+y=0\)
Vậy B=0