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lời giải
a)
\(\left(x+1\right)\left(2x-1\right)+x\le2x^2+3\)
\(\Leftrightarrow2x^2+x-1+x\le2x^2+3\)
\(\Leftrightarrow2x\le4\Rightarrow x\le2\)
\(\)b) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)-x>x^3+6x^2-5\)
\(\left(x^2+3x+2\right)\left(x+3\right)-x>x^3+6x^2-5\)
\(x^3+3x^2+3x^2+9x+2x+6-x>x^3+6x^2-5\)
\(10x+6>-5\Rightarrow x>-\dfrac{11}{10}\)
c)Đkxđ: x≥0
x+√x>(2√x+3)(√x−1)
⇔x+√x>2x+√x−3
⇔x−3>0
⇔x>3. (tmđk).
\(\begin{cases}xy\left(x+1\right)=x^3+y^2+x-y\left(1\right)\\3y\left(2+\sqrt{9x^2+3}\right)+\left(4y+2\right)\left(\sqrt{1+x+x^2}+1\right)=0\left(2\right)\end{cases}\)
Điều kiện xác định : mọi \(x\in Z\)
Ta có : \(xy\left(x+1\right)=x^3+y^2+x-y\Leftrightarrow x^3-x^2y+y^2-xy+x-y=0\)
\(\Leftrightarrow\left(x-y\right)\left(x^2-y-1\right)=0\Leftrightarrow\begin{cases}y=x\\y=x^2+1\end{cases}\)
Với \(y=x^2+1\) thay vào phương trình (2) ta được :
\(3\left(x^2+1\right)\left(2+\sqrt{9x^2+3}\right)+\left(4x^2+6\right)\left(\sqrt{1+x+x^2}+1\right)=0\)
Giải ra ta có phương trình vô nghiệm
Với y=x, thay vào phương trình thứ 2, ta được :
\(3x\left(2+\sqrt{9x^2+3}\right)+\left(4x+2\right)\left(\sqrt{1+x+x^2}+1\right)=0\)
\(\Leftrightarrow3x\left(2+\sqrt{9x^2+3}\right)=-\left(2x+1\right)\left(\sqrt{3+\left(2x+1\right)^2}+2\right)\)
\(\Leftrightarrow3x\left(2+\sqrt{9x^2+3}\right)=\left(-2x-1\right)\left(\sqrt{3+\left(-2x-1\right)^2}+2\right)\)
Xét hàm số \(f\left(t\right)=t\left(\sqrt{t^2+2}+2\right)\)
Ta có : \(f'\left(t\right)=\sqrt{t^2+2}+2+\frac{t^2}{\sqrt{t^2+2}}>0\) suy ra hàm số đồng biến
Từ đó suy ra \(3x=-2x\Leftrightarrow x=-\frac{1}{5}\)
Vậy hệ phương trình có nghiệm \(\left(x,y\right)=\left(-\frac{1}{5};-\frac{1}{5}\right)\)
8.
ĐKXĐ: \(x\ge\frac{2}{3}\)
\(\Leftrightarrow\frac{9\left(x+3\right)}{\sqrt{4x+1}+\sqrt{3x-2}}=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(l\right)\\\frac{9}{\sqrt{4x+1}+\sqrt{3x-2}}=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{4x+1}+\sqrt{3x-2}=9\)
\(\Leftrightarrow\sqrt{4x+1}-5+\sqrt{3x-2}-4=0\)
\(\Leftrightarrow\frac{4\left(x-6\right)}{\sqrt{4x+1}+5}+\frac{3\left(x-6\right)}{\sqrt{3x-2}+4}=0\)
\(\Leftrightarrow\left(x-6\right)\left(\frac{4}{\sqrt{4x+1}+5}+\frac{3}{\sqrt{3x-2}+4}\right)=0\)
\(\Leftrightarrow x=6\)
6.
ĐKXD: ...
\(\Leftrightarrow2\left(x^2-6x+9\right)+\left(x+5-4\sqrt{x+1}\right)=0\)
\(\Leftrightarrow2\left(x-3\right)^2+\frac{\left(x-3\right)^2}{x+5+4\sqrt{x+1}}=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(2+\frac{1}{x+5+4\sqrt{x+1}}\right)=0\)
\(\Leftrightarrow x=3\)
7.
\(\sqrt{x-\frac{1}{x}}-\sqrt{2x-\frac{5}{x}}+\frac{4}{x}-x=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x-\frac{1}{x}}=a\ge0\\\sqrt{2x-\frac{5}{x}}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=\frac{4}{x}-x\)
\(\Rightarrow a-b+a^2-b^2=0\)
\(\Leftrightarrow\left(a-b\right)\left(a+b+1\right)=0\)
\(\Leftrightarrow a=b\Leftrightarrow x-\frac{1}{x}=2x-\frac{5}{x}\)
\(\Leftrightarrow x=\frac{4}{x}\Rightarrow x=\pm2\)
Thế nghiệm lại pt ban đầu để thử (hoặc là bạn tìm ĐKXĐ từ đầu)
a/ ĐKXĐ: \(x^2+2x-6\ge0\)
\(\Leftrightarrow x^2+2x-6+\left(x-2\right)\sqrt{x^2+2x-6}=0\)
\(\Leftrightarrow\sqrt{x^2+2x-6}\left(\sqrt{x^2+2x-6}+x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+2x-6}=0\left(1\right)\\\sqrt{x^2+2x-6}=2-x\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x^2+2x-6=0\Rightarrow x=-1\pm\sqrt{7}\)
\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}2-x\ge0\\x^2+2x-6=\left(2-x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le2\\6x=10\end{matrix}\right.\) \(\Rightarrow x=\frac{5}{3}\)
Câu b nhìn ko ra hướng, ko biết đề có nhầm đâu ko :(
c/ ĐKXĐ: \(\left[{}\begin{matrix}x\ge0\\x\le-1\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{\left(x^2+x\right)\left(x^2+x+2\right)}-\left(3-x\right)\sqrt{x^2+x}=0\)
\(\Leftrightarrow\sqrt{x^2+x}\left(\sqrt{x^2+x+2}-3+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x=0\left(1\right)\\\sqrt{x^2+x+2}=3-x\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}3-x\ge0\\x^2+x+2=\left(3-x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le3\\7x=7\end{matrix}\right.\) \(\Rightarrow x=1\)
d/
Ta có \(\sqrt{x^2+3x+4}=\sqrt{\left(x+\frac{3}{4}\right)^2+\frac{7}{4}}>1\)
\(\Rightarrow\sqrt{x^2+3x+4}-1>0\)
Nhân 2 vế của pt với \(\sqrt{x^2+3x+4}-1\)
\(\left(\sqrt{x^2+3x+4}-1\right)\left(x^2+3x+3\right)=3x\left(x^2+3x+3\right)\)
\(\Leftrightarrow\left(x^2+3x+3\right)\left(\sqrt{x^2+3x+4}-1-3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+3x+3=0\left(vn\right)\\\sqrt{x^2+3x+4}=3x+1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Rightarrow\left\{{}\begin{matrix}x\ge-\frac{1}{3}\\x^2+3x+4=\left(3x+1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow8x^2+3x-3=0\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{-3+\sqrt{105}}{6}\\x=\frac{-3-\sqrt{105}}{6}\left(l\right)\end{matrix}\right.\)
ĐKXĐ: ...
\(\Leftrightarrow\left(x-4\right)\left(x^2-3x-3\right)=\left(x-3\right)\left(x-2+5\sqrt{x-3}\right).\frac{\left(x-4\right)}{\sqrt{x-3}+1}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2-3x-3=\frac{\left(x-3\right)\left(x-2+5\sqrt{x-3}\right)}{\sqrt{x-3}+1}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left(x^2-3x-3\right)\sqrt{x-3}+x^2-3x-3=x^2-5x+6+\left(5x-15\right)\sqrt{x-3}\)
\(\Leftrightarrow\left(x^2-8x+12\right)\sqrt{x-3}+2x-9=0\)
\(\Leftrightarrow\left(x^2-8x+12\right)\left(\sqrt{x-3}-x+4\right)+x^3-12x^2+46x-57=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2-9x+19\right)-\frac{\left(x^2-8x+12\right)\left(x^2-9x+19\right)}{\sqrt{x-3}+x-4}=0\)
\(\Leftrightarrow\left(x^2-9x+19\right)\left(x-3-\frac{x^2-8x+12}{\sqrt{x-3}+x-4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-9x+19=0\Rightarrow x=...\\x-3=\frac{x^2-8x+12}{\sqrt{x-3}+x-4}\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\left(x-3\right)\sqrt{x-3}+x^2-7x+12=x^2-8x+12\)
\(\Leftrightarrow\left(x-3\right)\sqrt{x-3}=-x\) (vô nghiệm do \(x\ge3\) nên vế trái không âm, vế phải luôn âm)