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NQ
2
5 tháng 7 2021
a) Ta có: \(A=3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}+30\)
\(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+30\)
\(=14\sqrt{2x}+30\)
b) Ta có: \(B=4\sqrt{\dfrac{25x}{4}}-\dfrac{8}{3}\sqrt{\dfrac{9x}{4}}-\dfrac{4}{3x}\cdot\sqrt{\dfrac{9x^3}{64}}\)
\(=4\cdot\dfrac{5\sqrt{x}}{2}-\dfrac{8}{3}\cdot\dfrac{3\sqrt{x}}{2}-\dfrac{4}{3x}\cdot\dfrac{3x\sqrt{x}}{8}\)
\(=10\sqrt{x}-4\sqrt{x}-\dfrac{1}{2}\sqrt{x}\)
\(=\dfrac{11}{2}\sqrt{x}\)
c) Ta có: \(\dfrac{y}{2}+\dfrac{3}{4}\sqrt{9y^2-6y+1}-\dfrac{3}{2}\)
\(=\dfrac{1}{2}y+\dfrac{3}{4}\left(1-3y\right)-\dfrac{3}{2}\)
\(=\dfrac{1}{2}y+\dfrac{3}{4}-\dfrac{9}{4}y-\dfrac{3}{2}\)
\(=-\dfrac{7}{4}y-\dfrac{3}{4}\)
10 tháng 1
\(\left\{{}\begin{matrix}\left(x+1\right)^2-y^2+6y-9=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x+1\right)^2-\left(y^2-6y+9\right)=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x+1\right)^2-\left(y-3\right)^2=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x+1+y-3\right)\left(x+1-y+3\right)=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x+y-2\right)\left(x-y+4\right)=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)
TH1: x+y-2=0
=>x=-y+2
\(2x^2+y^2-6y-1=0\)
=>\(2\left(-y+2\right)^2+y^2-6y-1=0\)
=>\(2\left(y^2-4y+4\right)+y^2-6y-1=0\)
=>\(3y^2-14y+7=0\)
\(\Delta=\left(-14\right)^2-2\cdot3\cdot7=196-42=154>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}y=\dfrac{14-\sqrt{154}}{6}\\y=\dfrac{14+\sqrt{154}}{6}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-y+2=\dfrac{-2+\sqrt{154}}{6}\\x=\dfrac{-2-\sqrt{154}}{6}\end{matrix}\right.\)
TH2: x-y+4=0
=>x=y-4
\(2x^2+y^2-6y-1=0\)
=>\(2\left(y-4\right)^2+y^2-6y-1=0\)
=>\(2\left(y^2-8y+16\right)+y^2-6y-1=0\)
=>\(3y^2-22y+31=0\)
\(\Delta=\left(-22\right)^2-4\cdot3\cdot31=112>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left[{}\begin{matrix}y_1=\dfrac{22-\sqrt{112}}{2\cdot3}=\dfrac{11-\sqrt{28}}{3}\\y_2=\dfrac{22+\sqrt{112}}{2\cdot3}=\dfrac{11+\sqrt{28}}{3}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=y-4=\dfrac{11-\sqrt{28}}{3}-4=\dfrac{-1-\sqrt{28}}{3}\\x=y-4=\dfrac{11+\sqrt{28}}{3}-4=\dfrac{-1+\sqrt{28}}{3}\end{matrix}\right.\)
H9
HT.Phong (9A5)
CTVHS
10 tháng 1
\(\left\{{}\begin{matrix}\left(x+1\right)^2-y^2+6y-9=0\\2x^2+y^2-6y+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^2-\left(y^2-6y+9\right)=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^2-\left(y-3\right)^2=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y-2\right)\left(x-y+4\right)=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y-2=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\\\left\{{}\begin{matrix}x-y+4=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2-y\\2\cdot\left(2-y\right)^2+y^2-6y-1=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=y-4\\2\cdot\left(y-4\right)^2+y^2-6y-1=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2-y\\3y^2-14y+7=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=y-4\\3y^2-22y+31=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-\dfrac{1+2\sqrt{7}}{3}\\y=\dfrac{7+2\sqrt{7}}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{-1+2\sqrt{7}}{3}\\y=\dfrac{7-2\sqrt{7}}{3}\end{matrix}\right.\end{matrix}\right.\\\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{-1+2\sqrt{7}}{3}\\y=\dfrac{11+2\sqrt{7}}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-\dfrac{1+2\sqrt{7}}{3}\\y=\dfrac{11-2\sqrt{7}}{3}\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)
Vậy các cặp (x;y) thỏa mãn là: \(\left(-\dfrac{1+2\sqrt{7}}{3};\dfrac{7+2\sqrt{7}}{3}\right);\left(\dfrac{-1+2\sqrt{7}}{3};\dfrac{7-2\sqrt{7}}{3}\right);\left(\dfrac{-1+2\sqrt{7}}{3};\dfrac{11+2\sqrt{7}}{3}\right);\left(-\dfrac{1+2\sqrt{7}}{3};\dfrac{11-2\sqrt{7}}{3}\right)\)
2
18 tháng 4 2023
`a)|x-2|=2<=>[(x=4(ko t//m)),(x=0(t//m)):}`
Thay `x=0` vào `A` có: `A=[2\sqrt{0}-3]/[\sqrt{0}-2]=3/2`
`b)` Với `x >= 0,x ne 4` có:
`B=[2(\sqrt{x}-3)+\sqrt{x}(\sqrt{x}+3)-4\sqrt{x}]/[(\sqrt{x}+3)(\sqrt{x}-3)]`
`B=[2\sqrt{x}-6+x+3\sqrt{x}-4\sqrt{x}]/[(\sqrt{x}+3)(\sqrt{x}-3)]`
`B=[x+\sqrt{x}-6]/[(\sqrt{x}+3)(\sqrt{x}-3)]`
`B=[(\sqrt{x}+3)(\sqrt{x}-2)]/[(\sqrt{x}+3)(\sqrt{x}-3)]`
`B=[\sqrt{x}-2]/[\sqrt{x}-3]`
`c)` Với `x >= 0,x ne 4` có:
`C=A.B=[2\sqrt{x}-3]/[\sqrt{x}-2].[\sqrt{x}-2]/[\sqrt{x}-3]=[2\sqrt{x}-3]/[\sqrt{x}-3]`
Có: `C >= 1`
`<=>[2\sqrt{x}-3]/[\sqrt{x}-3] >= 1`
`<=>[2\sqrt{x}-3-\sqrt{x}+3]/[\sqrt{x}-3] >= 0`
`<=>[\sqrt{x}]/[\sqrt{x}-3] >= 0`
Vì `x >= 0=>\sqrt{x} >= 0`
`=>\sqrt{x}-3 > 0`
`<=>x > 9` (t/m đk)
1 tháng 1 2022
a: Thay x=16 vào A, ta được:
\(A=\dfrac{2\cdot4}{4+3}=\dfrac{8}{7}\)
AH
Akai Haruma
Giáo viên
5 tháng 4 2021
Lời giải:
$x^3-9y^2+9x-6y=1$
$\Leftrightarrow x^3+9x=9y^2+6y+1$
$\Leftrightarrow x(x^2+9)=(3y+1)^2$
Đặt $(x,x^2+9)=d$ thì suy ra $9\vdots d(*)$
$(3y+1)^2=x(x^2+9)\vdots d^2\Rightarrow 3y+1\vdots d$. Mà $(3y+1,3)=1$ nên $(3,d)=1(**)$
Từ $(*);(**)\Rightarrow d=1$, hay $x,x^2+9$ nguyên tố cùng nhau.
$\Rightarrow \frac{x}{x^2+9}$ là phấn số tối giản.