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Câu 1:
\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=3\)
\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=3\)(1)
Trường hợp 1: x<1
(1) trở thành 1-x+2-x=3
=>3-2x=3
=>x=0(nhận)
Trường hợp 2: 1<=x<2
(1) trở thành x-1+2-x=3
=>1=3(loại)
Trường hợp 3: x>=2
(1) trở thành x-1+x-2=3
=>2x-3=3
=>2x=6
hay x=3(nhận)
`A=sqrt{x-2}+sqrt{6-x}(2<=x<=6)`
Áp dụng BĐT `sqrtA+sqrtB>=sqrt{A+B}`
`=>A>=sqrt{x-2+6-x}=2`
Dấu "=" `<=>x=2` hoặc `x=6`
Áp dụng BĐT bunhia
`=>A<=sqrt{2(x-2+6-x)}=2sqrt2`
Dấu "=" `<=>x=4`
`C=sqrt{1+x}+sqrt{8-x}(-1<=x<=8)`
Áp dụng BĐT `sqrtA+sqrtB>=sqrt{A+B}`
`=>A>=sqrt{1+x+8-x}=3`
Dấu "=" `<=>x=-1` hoặc `x=8`
Áp dụng BĐT bunhia
`=>A<=sqrt{2(1+x+8-x)}=3sqrt2`
Dấu "=" `<=>x=7/2`
`D=2sqrt{x+5}+sqrt{1-2x}(-5<=x<=1/2)`
`=sqrt{4x+20}+sqrt{1-2x}`
Áp dụng BĐT `sqrtA+sqrtB>=sqrt{A+B}`
`=>D>=sqrt{4x+20+1-2x}=sqrt{2x+21}`
Mà `x>=-5`
`=>D>=sqrt{-10+21}=sqrt{11}`
Dấu "=" `<=>x=-5`
\(A=x+\sqrt{x}\) có điều kiện xác định là: \(x\ge0\)
\(\Rightarrow A_{min}=0\) khi x = 0
\(B=x+5\sqrt{x+7}\) có điều kiện xác định là: \(x\ge-7\)
\(\Rightarrow B_{min}=-7+5\cdot0=-7\) khi x = -7
\(C=2x-6\sqrt{x+1}\) có điều kiện xác định là \(x\ge-1\)
\(\Rightarrow C_{min}=2\cdot\left(-1\right)-6\cdot0=-2\) khi x = -1
a: ĐKXĐ: x>=0; x<>1
b \(A=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\)
\(=\dfrac{x+2\sqrt{x}-x-\sqrt{x}-1}{x\sqrt{x}-1}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\dfrac{1}{\sqrt{x}+2}\)
c: Khi x=9-4 căn 5 thì \(A=\dfrac{1}{\sqrt{5}-2+2}=\dfrac{\sqrt{5}}{5}\)
d: căn x+2>=2
=>A<=1/2
Dấu = xảy ra khi x=0
a) \(\sqrt[]{x^2-4x+4}=x+3\)
\(\Leftrightarrow\sqrt[]{\left(x-2\right)^2}=x+3\)
\(\Leftrightarrow\left|x-2\right|=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\\x-2=-\left(x+3\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}0x=5\left(loại\right)\\x-2=-x-3\end{matrix}\right.\)
\(\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)
b) \(2x^2-\sqrt[]{9x^2-6x+1}=5\)
\(\Leftrightarrow2x^2-\sqrt[]{\left(3x-1\right)^2}=5\)
\(\Leftrightarrow2x^2-\left|3x-1\right|=5\)
\(\Leftrightarrow\left|3x-1\right|=2x^2-5\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2x^2-5\\3x-1=-2x^2+5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-4=0\left(1\right)\\2x^2+3x-6=0\left(2\right)\end{matrix}\right.\)
Giải pt (1)
\(\Delta=9+32=41>0\)
Pt \(\left(1\right)\) \(\Leftrightarrow x=\dfrac{3\pm\sqrt[]{41}}{4}\)
Giải pt (2)
\(\Delta=9+48=57>0\)
Pt \(\left(2\right)\) \(\Leftrightarrow x=\dfrac{-3\pm\sqrt[]{57}}{4}\)
Vậy nghiệm pt là \(\left[{}\begin{matrix}x=\dfrac{3\pm\sqrt[]{41}}{4}\\x=\dfrac{-3\pm\sqrt[]{57}}{4}\end{matrix}\right.\)
\(a_1,\sqrt{x}< 7\\ \Rightarrow x< 49\\ a_2,\sqrt{2x}< 6\\ \Rightarrow x< 18\\ a_3,\sqrt{4x}\ge4\\ \Rightarrow4x\ge16\\ \Rightarrow x\ge4\\ a_4,\sqrt{x}< \sqrt{6}\\ \Rightarrow x< 6\)
\(b_1,\sqrt{x}>4\\ \Rightarrow x>16\\ b_2,\sqrt{2x}\le2\\ \Rightarrow2x\le4\\ \Rightarrow x\le2\\ b_3,\sqrt{3x}\le\sqrt{9}\\ \Rightarrow3x\le9\\ \Rightarrow x\le3\\ b_4,\sqrt{7x}\le\sqrt{35}\\ \Rightarrow7x\le35\\ \Rightarrow x\le5\)
a, \(A=\left(\frac{1}{1-\sqrt{x}}+\frac{1}{1+\sqrt{x}}\right):\left(\frac{1}{1-\sqrt{x}}-\frac{1}{1+\sqrt{x}}\right)+\frac{1}{1-\sqrt{x}}\)ĐK : \(x>0;x\ne1\)
\(=\left(\frac{1+\sqrt{x}+1-\sqrt{x}}{1-x}\right):\left(\frac{1+\sqrt{x}-1+\sqrt{x}}{1-x}\right)+\frac{1}{1-\sqrt{x}}\)
\(=\frac{2}{1-x}.\frac{1-x}{2\sqrt{x}}+\frac{1}{1-\sqrt{x}}=\frac{1}{\sqrt{x}}+\frac{1}{1-\sqrt{x}}=\frac{1-\sqrt{x}+\sqrt{x}}{-x+\sqrt{x}}=\frac{1}{\sqrt{x}-x}\)
b, Ta có : \(x=7+4\sqrt{3}=7+2.2\sqrt{3}=\left(\sqrt{4}+\sqrt{3}\right)^2\)
\(A=\frac{1}{\sqrt{4}+\sqrt{3}-7+4\sqrt{3}}\)
1)\(A=\sqrt{x^2-2x+1}+\sqrt{x^2+2x+1}\\ A=\left|x-1\right|+\left|x+1\right|\\ A=\left|1-x\right|+\left|x+1\right|\ge\left|1-x+x+1\right|=2\)
dấu "=" xảy ra khi \(\left[{}\begin{matrix}\left\{{}\begin{matrix}1-x\ge0\\x+1\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}1-x< 0\\x+1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}1\ge x\\x\ge-1\end{matrix}\right.\left(nhận\right)\\\left\{{}\begin{matrix}1< x\\x< -1\end{matrix}\right.\left(loại\right)\end{matrix}\right.\)
vậy....
\(B=\sqrt{4x^2-12x+9}+\sqrt{4x^2+12x+9}\\ B=\left|2x-3\right|+\left|2x+3\right|\\ B=\left|3-2x\right|+\left|2x+3\right|\ge\left|3-2x+2x+3\right|=6\)
dấu " = " xảy ra khi \(\left[{}\begin{matrix}\left\{{}\begin{matrix}3-2x\ge0\\2x+3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}3-2x< 0\\2x+3< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3\ge2x\\2x\ge-3\end{matrix}\right.\\\left\{{}\begin{matrix}3< 2x\\2x< -3\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\dfrac{3}{2}\ge x\\x\ge-\dfrac{3}{2}\end{matrix}\right.\left(nhận\right)\\\left\{{}\begin{matrix}\dfrac{3}{2}< x\\x< -\dfrac{3}{2}\end{matrix}\right.\left(loại\right)\end{matrix}\right.\)
vậy....
2)
\(A=\sqrt{x+4}+\sqrt{4-x}\\ A^2=x+4+4-x+2\sqrt{\left(x+4\right)\left(4-x\right)}\\ A^2=4+2\sqrt{16-x^2}\\ vìx^2\ge0nên\\ A^2\le12\\ A\le\sqrt{12}\)
dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x^2\ge0\\x^2\le16\end{matrix}\right.\Rightarrow0\le x\le4\)
vậy...
\(B=\sqrt{x+6}+\sqrt{6-x}\\ B^2=x+6+6-x+2\sqrt{\left(x+6\right)\left(6-x\right)}\\ B^2=12+2\sqrt{36-x^2}\\ vì\: x^2\ge0nên\\ B^2\le24\\ B\le\sqrt{24}\)
dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x^2\ge0\\x^2\le36\end{matrix}\right.\Rightarrow0\le x\le6\)
bạn có cách nào làm cho x nó ra 1 số cụ thể ko ??