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1. \(\Leftrightarrow\sqrt{\left(\sqrt{x}-2\right)^2}+\sqrt{\left(\sqrt{x}-3\right)^2}=1\)
\(\Leftrightarrow\left|\sqrt{x}-2\right|+\left|3-\sqrt{x}\right|=1\)
+ Ta có : \(\left|\sqrt{x}-2\right|+\left|3-\sqrt{x}\right|\ge\left|\sqrt{x}-2+3-\sqrt{x}\right|=1\)
Dấu "=" \(\Leftrightarrow\left(\sqrt{x}-2\right)\left(3-\sqrt{x}\right)\ge0\)
\(\Leftrightarrow2\le\sqrt{x}\le3\Leftrightarrow4\le x\le9\)
2. + \(ĐK:4-2x-x^2\ge0\)
+ VT = \(\sqrt{3\left(x^2+2x+1\right)+4}+\sqrt{5\left(x^2+2x+1\right)+9}\)
\(=\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+9}\) \(\ge\sqrt{4}+\sqrt{9}=5\) (1)
Dấu "=" \(\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)
+ VP \(=-\left(x^2+2x+1\right)+5=-\left(x+1\right)^2+5\le5\forall x\) (2)
Dấu "=" \(\Leftrightarrow x=-1\)
+ Từ (1) và (2) suy ra : pt \(\Leftrightarrow VT=VP=5\Leftrightarrow x=-1\) (TM)
3. + TH1: \(x< 0\) ta có :
\(VT< \sqrt[3]{2.0+1}+\sqrt[3]{0}=1\) ( KTM )
+ TH2 : x = 0 ta có :
\(VT=\sqrt[3]{1}+\sqrt[3]{0}=1\) ( TM )
+ TH3 : x > 0 ta có :
\(VT>\sqrt[3]{2.0+1}+\sqrt[3]{0}=1\) ( KTM )
Vậy x = 0 là nghiệm duy nhất của pt
4. \(\Leftrightarrow\left(x-1\right)\left(x+4\right)\left(x-2\right)\left(x+3\right)-24=0\)
\(\Leftrightarrow\left(x^2+2x-3\right)\left(x^2+2x-8\right)-24=0\)
\(\Leftrightarrow t\left(t-5\right)-24=0\) ( với \(t=x^2+2x-3\) )
\(\Leftrightarrow t^2-5t-24=0\Leftrightarrow\left(t+3\right)\left(t-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-3\\t=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+2x-3=-3\\x^2+2x-3=8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(x+2\right)=0\\\left(x+1\right)^2=12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=2\sqrt{3}-1\\x=-2\sqrt{3}-1\end{matrix}\right.\) ( TM )
1)
ĐK: \(x\geq 5\)
PT \(\Leftrightarrow \sqrt{4(x-5)}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9(x-5)}=6\)
\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=6\)
\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=6\)
\(\Leftrightarrow 2\sqrt{x-5}=6\Rightarrow \sqrt{x-5}=3\Rightarrow x=3^2+5=14\)
2)
ĐK: \(x\geq -1\)
\(\sqrt{x+1}+\sqrt{x+6}=5\)
\(\Leftrightarrow (\sqrt{x+1}-2)+(\sqrt{x+6}-3)=0\)
\(\Leftrightarrow \frac{x+1-2^2}{\sqrt{x+1}+2}+\frac{x+6-3^2}{\sqrt{x+6}+3}=0\)
\(\Leftrightarrow \frac{x-3}{\sqrt{x+1}+2}+\frac{x-3}{\sqrt{x+6}+3}=0\)
\(\Leftrightarrow (x-3)\left(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}\right)=0\)
Vì \(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}>0, \forall x\geq -1\) nên $x-3=0$
\(\Rightarrow x=3\) (thỏa mãn)
Vậy .............
1/ Đặt \(\sqrt{x^2+2}=t>0\Rightarrow x^2=t^2-2\)
\(t^2-2+\left(3-t\right)x-1-2t=0\)
\(\Leftrightarrow t^2-2t-3-\left(t-3\right)x=0\)
\(\Leftrightarrow\left(t-3\right)\left(t+1\right)-\left(t-3\right)x=0\)
\(\Leftrightarrow\left(t-3\right)\left(t+1-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t-3=0\\t+1-x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}t=3\\t=x-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+2}=3\left(1\right)\\\sqrt{x^2+2}=x-1\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x^2=7\Rightarrow x=\pm\sqrt{7}\)
\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x^2+2=\left(x-1\right)^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x^2+2=x^2-2x+1\end{matrix}\right.\) \(\Rightarrow x=\dfrac{-1}{2}\left(l\right)\)
Vậy nghiệm pt là \(x=\pm\sqrt{7}\)
2/
\(x^2+3-6x\sqrt{x^2+3}+9x^2-\sqrt{x^2+3}+3x-2=0\)
\(\Leftrightarrow\left(\sqrt{x^2+3}-3x\right)^2-\left(\sqrt{x^2+3}-3x\right)-2=0\)
Đặt \(\sqrt{x^2+3}-3x=t\)
\(\Rightarrow t^2-t-2=0\) \(\Rightarrow\left[{}\begin{matrix}t=-1\\t=2\end{matrix}\right.\)
TH1: \(\sqrt{x^2+3}-3x=-1\Rightarrow\sqrt{x^2+3}=3x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-1\ge0\\x^2+3=\left(3x-1\right)^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\8x^2-6x-2=0\end{matrix}\right.\) \(\Rightarrow x=1\)
TH2: \(\sqrt{x^2+3}-3x=2\Leftrightarrow\sqrt{x^2+3}=3x+2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{-2}{3}\\x^2+3=\left(3x+2\right)^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{-2}{3}\\8x^2+12x+1=0\end{matrix}\right.\) \(\Rightarrow x=\dfrac{-3+\sqrt{7}}{4}\)
3/ ĐKXĐ: \(\dfrac{3}{2}\le x\le\dfrac{5}{2}\)
\(1.\sqrt{2x-3}+1.\sqrt{5-2x}\le\sqrt{\left(1^2+1^2\right)\left(2x-3+5-2x\right)}=2\)
\(\Rightarrow VT\le2\)
\(VP=3\left(x^2-4x+4\right)+2=3\left(x-2\right)^2+2\ge2\)
\(\Rightarrow VT=VP\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\2x-3=5-2x\end{matrix}\right.\) \(\Rightarrow x=2\)
Vậy pt có nghiệm duy nhất \(x=2\)
4/
ĐKXĐ: \(x\ge\dfrac{-5}{4}\)
\(x^2-2x+1+4x+5-6\sqrt{4x+5}+9=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(\sqrt{4x+5}-3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\\sqrt{4x+5}-3=0\end{matrix}\right.\) \(\Rightarrow x=1\)
Vậy pt có nghiệm duy nhất \(x=1\)
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!
Sorry nha , em ko bt làm đâu , em mới học lớp 5 thui
sory nha ae cũng ko biết làm đâu... em mới lên lớp 6 thôi