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a) ĐKXĐ: \(3\le x\le10\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}x>-4\\x\ne4\end{matrix}\right.\)
c) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\x\ne4\end{matrix}\right.\)
d) ĐKXĐ: \(x\ge\dfrac{1}{2}\)
e) ĐKXĐ: \(x\in R\)
Câu a:
ĐKXĐ:...........
\(\sqrt{x^2-x+9}=2x+1\)
\(\Rightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-x+9=(2x+1)^2=4x^2+4x+1\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+5x-8=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x(x-1)+8(x-1)=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ (x-1)(3x+8)=0\end{matrix}\right.\Rightarrow x=1\)
Vậy.....
Câu b:
ĐKXĐ:.........
Ta có: \(\sqrt{5x+7}-\sqrt{x+3}=\sqrt{3x+1}\)
\(\Rightarrow (\sqrt{5x+7}-\sqrt{x+3})^2=3x+1\)
\(\Leftrightarrow 5x+7+x+3-2\sqrt{(5x+7)(x+3)}=3x+1\)
\(\Leftrightarrow 3(x+3)=2\sqrt{(5x+7)(x+3)}\)
\(\Leftrightarrow \sqrt{x+3}(3\sqrt{x+3}-2\sqrt{5x+7})=0\)
Vì \(x\geq -\frac{7}{5}\Rightarrow \sqrt{x+3}>0\). Do đó:
\(3\sqrt{x+3}-2\sqrt{5x+7}=0\)
\(\Rightarrow 9(x+3)=4(5x+7)\)
\(\Rightarrow 11x=-1\Rightarrow x=\frac{-1}{11}\) (thỏa mãn)
Vậy..........
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
a) + \(VT=\sqrt{x^2+2x+10}+x^2+2x+1+7\)
\(=\sqrt{x^2+2x+1}+\left(x+1\right)^2+7>0\forall x\)
=> ptvn
d) ĐK : \(x^2+7x+7\ge0\)
Đặt \(t=\sqrt{x^2+7x+7}\ge0\) \(\Rightarrow t^2=x^2+7x+7\)
\(pt\Leftrightarrow3\left(x^2+7x+7\right)-3+2\sqrt{x^2+7x+7}-2=0\)
\(\Leftrightarrow3t^2+2t-5=0\Leftrightarrow\left(3t+5\right)\left(t-1\right)=0\)
\(\Leftrightarrow t=1\) ( do \(3t+5>0\forall t\ge0\) )
\(\Leftrightarrow x^2+7x+1=0\Leftrightarrow x^2+7x+6=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\) ( TM )
f) ĐK : \(x\ge1\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-1}\ge0\\b=\sqrt{x+3}\ge0\end{matrix}\right.\) thì pt trở thành :
\(a+b-ab-1=0\)
\(\Leftrightarrow\left(a-1\right)-b\left(a-1\right)=0\)
\(\Leftrightarrow\left(1-b\right)\left(a-1\right)=0\Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x+3}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\x=-2\left(KTM\right)\end{matrix}\right.\)
a) \(\text{Đ}K\text{X}\text{Đ}:\frac{3}{2}\le x\le\frac{5}{2}\)
Áp dụng BĐT Bunhiacopxki ta có:
\(VT=\sqrt{2x-3}+\sqrt{5-2x}\le\sqrt{2\left(2x-3+5-2x\right)}=2\)
Dấu '=' xảy ra khi \(\sqrt{2x-3}=\sqrt{5-2x}\Leftrightarrow x=2\)
Lại có: \(VP=3x^2-12x+14=3\left(x-2\right)^2+2\ge2\)
Dấu '=' xảy ra khi x=2
Do đó VT=VP khi x=2
b) ĐK: \(x\ge0\). Ta thấy x=0 k pk là nghiệm của pt, chia 2 vế cho x ta có:
\(x^2-2x-x\sqrt{x}-2\sqrt{x}+4=0\Leftrightarrow x-2-\sqrt{x}-\frac{2}{\sqrt{x}}+\frac{4}{x}=0\)
\(\Leftrightarrow\left(x+\frac{4}{x}\right)-\left(\sqrt{x}+\frac{2}{\sqrt{x}}\right)-2=0\)
Đặt \(\sqrt{x}+\frac{2}{\sqrt{x}}=t>0\Leftrightarrow t^2=x+4+\frac{4}{x}\Leftrightarrow x+\frac{4}{x}=t^2-4\), thay vào ta có:
\(\left(t^2-4\right)-t-2=0\Leftrightarrow t^2-t-6=0\Leftrightarrow\left(t-3\right)\left(t+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}t=3\\t=-2\end{cases}}\)
Đối chiếu ĐK của t
\(\Rightarrow t=3\Leftrightarrow\sqrt{x}+\frac{2}{\sqrt{x}}=3\Leftrightarrow x-3\sqrt{x}+2=0\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=1\end{cases}}\)
\(a,\sqrt{x^4}=7\Leftrightarrow x^2=7\Leftrightarrow x=\pm\sqrt{7}\)
\(Dk:x\ge\frac{2}{3};\sqrt{3x-2}=4\Leftrightarrow3x-2=16\Leftrightarrow3x=18\Leftrightarrow x=6\left(tm\right)\)
\(dk:x\ge\frac{3}{2};\sqrt{2x-3}=\sqrt{x-1}\Leftrightarrow2x-3=x-1\Leftrightarrow x=2\left(tm\right)\)
\(dk:x\ge0;x-10\sqrt{x}+25=0\Leftrightarrow\left(\sqrt{x}-5\right)^2=0\Leftrightarrow\sqrt{x}=5\Leftrightarrow x=25\left(tm\right)\)
\(\sqrt{2x}< 3\Leftrightarrow\sqrt{2}.\sqrt{x}< 3\Leftrightarrow0\le\sqrt{x}< \sqrt{4,5}\Leftrightarrow0\le x< 4,5\)
\(h,dk:x\ge3;\sqrt{\left(x-1\right)^2}=3x-9\Leftrightarrow\left|x-1\right|=3x-9\Leftrightarrow x-1=3x-9\left(x\ge3\right)\Leftrightarrow x=4\left(tm\right)\)