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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).
a/ ĐKXĐ: \(-2\le x\le5\)
\(\sqrt{x+2}+\sqrt{5-x}+\sqrt{\left(x+2\right)\left(5-x\right)}-4=0\)
Đặt \(\sqrt{x+2}+\sqrt{5-x}=a>0\Rightarrow\sqrt{\left(x+2\right)\left(5-x\right)}=\frac{a^2-7}{2}\)
\(\Rightarrow a+\frac{a^2-7}{2}-4=0\)
\(\Leftrightarrow a^2+2a-15=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-5\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{\left(x+2\right)\left(5-x\right)}=\frac{a^2-7}{2}=1\)
\(\Leftrightarrow-x^2+3x+10=1\)
\(\Leftrightarrow x^2-3x-9=0\)
b/ \(\Leftrightarrow\sqrt{x+1}-\sqrt{4-x}+2\left(5+2\sqrt{\left(x+1\right)\left(4-x\right)}\right)=17\)
Đặt \(\sqrt{x+1}-\sqrt{4-x}=a\Rightarrow\sqrt{\left(x+1\right)\left(4-x\right)}=\frac{5-a^2}{2}\)
\(a+2\left(5+5-a^2\right)=17\)
\(\Leftrightarrow-2a^2+a+3=0\Rightarrow\left[{}\begin{matrix}a=-1\\a=\frac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x+1}-\sqrt{4-x}=-1\\\sqrt{x+1}-\sqrt{4-x}=\frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}+1=\sqrt{4-x}\\2\sqrt{x+1}=2\sqrt{4-x}+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2+2\sqrt{x+1}=4-x\\4x+4=25-4x+12\sqrt{4-x}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=1-x\left(x\le1\right)\\12\sqrt{4-x}=8x-21\left(x\ge\frac{21}{8}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=\left(1-x\right)^2\\144\left(4-x\right)=\left(8x-21\right)^2\end{matrix}\right.\)
c/ ĐKXĐ: \(0\le x\le1\)
Đặt \(\sqrt{x}+\sqrt{1-x}=a>0\Rightarrow\sqrt{x-x^2}=\frac{a^2-1}{2}\)
\(a^2-1=3\left(a-1\right)\Leftrightarrow a^2-3a+2=0\Rightarrow\left[{}\begin{matrix}a=1\\a=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x-x^2}=\frac{a^2-1}{2}=0\\\sqrt{x-x^2}=\frac{a^2-1}{2}=\frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-x^2=0\\x-x^2=\frac{9}{4}\left(vn\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
d/ ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{5+2x}=a\ge0\\\sqrt{5-2x}=b\ge0\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}\left(3a-1\right)\left(3b-1\right)=16\\a^2+b^2=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3ab-\left(a+b\right)=5\\\left(a+b\right)^2-2ab=10\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=3ab-5\\\left(a+b\right)^2-2ab=10\end{matrix}\right.\)
\(\Rightarrow\left(3ab-5\right)^2-2ab=10\)
\(\Leftrightarrow9\left(ab\right)^2-32ab+15=0\Rightarrow\left[{}\begin{matrix}ab=3\\ab=\frac{5}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left(ab\right)^2=9\\\left(ab\right)^2=\frac{25}{81}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}25-4x^2=9\\25-4x^2=\frac{25}{81}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=4\\x^2=\frac{500}{81}\end{matrix}\right.\)
\(\sqrt{2-f\left(x\right)}=f\left(x\right)\Leftrightarrow\left\{{}\begin{matrix}f\left(x\right)\ge0\\f^2\left(x\right)+f\left(x\right)-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}f\left(x\right)=1\\f\left(x\right)=-2< 0\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow f\left(1\right)=f\left(2\right)=f\left(3\right)=1\)
\(\sqrt{2g\left(x\right)-1}+\sqrt[3]{3g\left(x\right)-2}=2.g\left(x\right)\)
\(VT=1.\sqrt{2g\left(x\right)-1}+1.1\sqrt[3]{3g\left(x\right)-2}\)
\(VT\le\dfrac{1}{2}\left(1+2g\left(x\right)-1\right)+\dfrac{1}{3}\left(1+1+3g\left(x\right)-2\right)\)
\(\Leftrightarrow VT\le2g\left(x\right)\)
Dấu "=" xảy ra khi và chỉ khi \(g\left(x\right)=1\)
\(\Rightarrow g\left(0\right)=g\left(3\right)=g\left(4\right)=g\left(5\right)=1\)
Để các căn thức xác định \(\Rightarrow\left\{{}\begin{matrix}f\left(x\right)-1\ge0\\g\left(x\right)-1\ge0\end{matrix}\right.\)
Ta có:
\(\sqrt{f\left(x\right)-1}+\sqrt{g\left(x\right)-1}+f\left(x\right).g\left(x\right)-f\left(x\right)-g\left(x\right)+1=0\)
\(\Leftrightarrow\sqrt{f\left(x\right)-1}+\sqrt{g\left(x\right)-1}+\left[f\left(x\right)-1\right]\left[g\left(x\right)-1\right]=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}f\left(x\right)=1\\g\left(x\right)=1\end{matrix}\right.\) \(\Leftrightarrow x=3\)
Vậy tập nghiệm của pt đã cho có đúng 1 phần tử
\(\Leftrightarrow2m.2^x+\left(2m+1\right)\left(3-\sqrt{5}\right)^x+\left(3+\sqrt{5}\right)^x=0\)
\(\Leftrightarrow\left(\frac{3+\sqrt{5}}{2}\right)^x+\left(2m+1\right)\left(\frac{3-\sqrt{5}}{2}\right)^x+2m< 0\)
Đặt \(t=\left(\frac{3+\sqrt{5}}{2}\right)^x,0< t\le1\Rightarrow\frac{1}{t}=\left(\frac{3-\sqrt{5}}{2}\right)^x\)
Phương trình trở thành :
\(t+\left(2m+1\right)\frac{1}{t}+2m=0\) (*)
a. Khi \(m=-\frac{1}{2}\) ta có \(t=1\) suy ra \(\left(\frac{3+\sqrt{5}}{2}\right)^x=1\Leftrightarrow x=0\)
Vậy phương trình có nghiệm là \(x=0\)
b. Phương trình (*) \(\Leftrightarrow t^2+1=-2m\left(t+1\right)\Leftrightarrow\frac{t^2+1}{t+1}=-2m\)
Xét hàm số \(f\left(t\right)=\frac{t^2+1}{t+1};t\in\)(0;1]
Ta có : \(f'\left(t\right)=\frac{t^2+2t+1}{\left(t+1\right)^2}\Rightarrow f'\left(t\right)=0\Leftrightarrow=-1+\sqrt{2}\)
t f'(t) f(t) 0 1 0 - + 1 1 -1 + căn 2 2 căn 2 - 2
Suy ra phương trình đã cho có nghiệm đúng
\(\Leftrightarrow2\sqrt{2}-2\le-2m\le1\Leftrightarrow\sqrt{2}-1\ge m\ge-\frac{1}{2}\)
Vậy \(m\in\left[-\frac{1}{2};\sqrt{2}-1\right]\) là giá trị cần tìm
câu 4 \(\sqrt{x^2-2x}=\sqrt{2x-x^2}\Leftrightarrow x^2-2x=2x-x^2\)
\(\Leftrightarrow2\left(x^2-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
câu C
Câu 5 \(x\left(x^2-1\right)\sqrt{x-1}=0\)
ĐK \(x\ge1\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\sqrt{x-1}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\\sqrt{x-1}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nh\right)\\x=-1\left(l\right)\end{matrix}\right.\)
vậy pt có 1 nghiệm
câu B
e/ ĐKXĐ: \(-1\le x\le4\)
Tưởng nó giống câu c mà ko phải
\(\Leftrightarrow\sqrt{x+1}+\sqrt{4-x}+\sqrt{\left(4-x\right)\left(x+1\right)}=5\)
Đặt \(\sqrt{x+1}+\sqrt{4-x}=a>0\Rightarrow a^2=5+2\sqrt{\left(x+1\right)\left(4-x\right)}\)
\(\Rightarrow\sqrt{\left(x+1\right)\left(4-x\right)}=\frac{a^2-5}{2}\) pt trở thành:
\(a+\frac{a^2-5}{2}=5\Leftrightarrow a^2+2a-15=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-5\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x+1}+\sqrt{4-x}=3\)
\(\Leftrightarrow5+2\sqrt{-x^2+3x+4}=9\)
\(\Leftrightarrow\sqrt{-x^2+3x+4}=2\)
\(\Leftrightarrow-x^2+3x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
b/ĐKXĐ: \(0\le x\le4\)
\(\Leftrightarrow\left(3x-7\right)\sqrt{x\left(4-x\right)}+4-x=0\)
\(\Leftrightarrow\sqrt{4-x}\left[\left(3x-7\right)\sqrt{x}+\sqrt{4-x}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\\sqrt{4-x}=\left(7-3x\right)\sqrt{x}\left(x\le\frac{7}{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow4-x=x\left(7-3x\right)^2\)
\(\Leftrightarrow4-x=x\left(9x^2-42x+49\right)\)
\(\Leftrightarrow9x^3-42x^2+50x-4=0\)
\(\Leftrightarrow\left(x-2\right)\left(9x^2-24x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{4+\sqrt{14}}{3}>\frac{7}{3}\left(l\right)\\x=\frac{4-\sqrt{14}}{3}\end{matrix}\right.\)
1/ Đặt \(\sqrt[3]{x^2+5x-2}=t\Rightarrow x^2+5x=t^3+2\)
\(t^3+2=2t-2\)
\(\Leftrightarrow t^3-2t+4=0\)
\(\Leftrightarrow\left(t+2\right)\left(t^2-2t+2\right)=0\)
\(\Rightarrow t=-2\)
\(\Rightarrow\sqrt[3]{x^2+5x-2}=-2\)
\(\Leftrightarrow x^2+5x-2=-8\)
\(\Leftrightarrow x^2+5x+6=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
2/ \(\Leftrightarrow2x+11+3\sqrt[3]{\left(x+5\right)\left(x+6\right)}\left(\sqrt[3]{x+5}+\sqrt[3]{x+6}\right)=2x+11\)
\(\Leftrightarrow\sqrt[3]{\left(x+5\right)\left(x+6\right)}\left(\sqrt[3]{x+5}+\sqrt[3]{x+6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt[3]{x+5}=0\\\sqrt[3]{x+6}=0\\\sqrt[3]{x+5}=-\sqrt[3]{x+6}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-6\\x+5=-x-6\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-5\\x=-6\\x=-\frac{11}{2}\end{matrix}\right.\)
ĐK: \(x - 1 \ge 0\,\, \Leftrightarrow \,\,x \ge 1\)
\( \Rightarrow \) TXĐ của phương trình là: \(D = \left[ {1; + \infty } \right)\)
Giải phương trình: \(\sqrt {2{x^2} - 3} = x - 1\)
\(\begin{array}{l} \Leftrightarrow \,\,{\left( {\sqrt {2{x^2} - 3} } \right)^2} = {\left( {x - 1} \right)^2}\\ \Leftrightarrow \,\,2{x^2} - 3 = {x^2} - 2x + 1\\ \Leftrightarrow \,\,{x^2} + 2x - 4 = 0\\ \Leftrightarrow \,\,\left[ {\begin{array}{*{20}{c}}{x = - 1 + \sqrt 5 }\\{x = - 1 - \sqrt 5 }\end{array}} \right.\end{array}\)
Ta thấy \(x = - 1 + \sqrt 5 \) thỏa mãn.
Vậy tập nghiệm của phương trình là: \(S = \left\{ { - 1 + \sqrt 5 } \right\}\)
Chọn C.