Câu 1. Giải các phương trình, hệ phương trình sau:
a. \(\sqrt{x^2+x+4}+\sqrt{x^2+x+1}=\sqrt{2x^2+2x+9}\)
b. \(\left\{{}\begin{matrix}x^2-xy-2=0\\x^2+y^2+2x+2y-2=0\end{matrix}\right.\) (x,y \(\in R\))
a) Giải bất phương trình:
\(\sqrt{x^2+2x}+\sqrt{x^2+3x}\) ≥ \(2x\)
b) Giải hệ phương trình
\(\left\{{}\begin{matrix}x^3+6x^2y+9xy^2+y^3=0\\\sqrt{x-y}+\sqrt{x+y}=2\end{matrix}\right.\)
a, ĐKXĐ : \(\left[{}\begin{matrix}x\le-3\\x\ge0\end{matrix}\right.\)
TH1 : \(x\le-3\) ( LĐ )
TH2 : \(x\ge0\)
BPT \(\Leftrightarrow x^2+2x+x^2+3x+2\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge4x^2\)
\(\Leftrightarrow\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge x^2-\dfrac{5}{2}x\)
\(\Leftrightarrow2\sqrt{\left(x+2\right)\left(x+3\right)}\ge2x-5\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\x\ge-2\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\4x^2+20x+24\ge4x^2-20x+25\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}0\le x< \dfrac{5}{2}\\x\ge\dfrac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow x\ge0\)
Vậy \(S=R/\left(-3;0\right)\)
Giải các hệ phương trình sau:a) \(\left\{{}\begin{matrix}\left(2x-y\right)^2-6x+3y=0\\x+2y=0\end{matrix}\right.\);b) \(\left\{{}\begin{matrix}\sqrt{\dfrac{2x-y}{x+y}}+\sqrt{\dfrac{x+y}{2x-y}}=2\\3x+y=14\end{matrix}\right.\)
a.
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)^2-3\left(2x-y\right)=0\\x+2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)\left(2x-y-3\right)=0\\x+2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-y=0\\x+2y=0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-y-3=0\\x+2y=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{6}{5}\\y=-\dfrac{3}{5}\end{matrix}\right.\end{matrix}\right.\)
b.
ĐKXĐ: \(\dfrac{2x-y}{x+y}>0\)
Đặt \(\sqrt{\dfrac{2x-y}{x+y}}=t>0\) pt đầu trở thành:
\(t+\dfrac{1}{t}=2\Leftrightarrow t^2-2t+1=0\)
\(\Leftrightarrow t=1\Leftrightarrow\sqrt{\dfrac{2x-y}{x+y}}=1\)
\(\Leftrightarrow2x-y=x+y\Leftrightarrow x=2y\)
Thay xuống pt dưới:
\(6y+y=14\Rightarrow y=2\)
\(\Rightarrow x=4\)
Giải phương trình và hệ phương trình:
1) \(-2x^2+x+1-2\sqrt{x^2+x+1}=0\)
2) \(\left\{{}\begin{matrix}x^4+y^3x+x^2y^2=3y^4\\2x^2+y^4+1=2x\left(y^2+1\right)\end{matrix}\right.\)
1) \(-2x^2+x+1-2\sqrt[]{x^2+x+1}=0\)
\(\Leftrightarrow2\sqrt[]{x^2+x+1}=-2x^2+x+1\left(1\right)\)
Ta có :
\(2\sqrt[]{x^2+x+1}=2\sqrt[]{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\ge\sqrt[]{3}\)
Dấu "=" xảy ra khi và chỉ khi \(x+\dfrac{1}{2}=0\Leftrightarrow x=-\dfrac{1}{2}\)
\(\left(1\right)\Leftrightarrow-2x^2+x+1=\sqrt[]{3}\)
\(\Leftrightarrow2x^2-x+\sqrt[]{3}-1=0\)
\(\Delta=1-8\left(\sqrt[]{3}-1\right)=9-8\sqrt[]{3}\)
\(pt\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt[]{9-8\sqrt[]{3}}}{4}\left(loại\right)\\x=\dfrac{1-\sqrt[]{9-8\sqrt[]{3}}}{4}\left(loại\right)\end{matrix}\right.\) \(\left(vì.x=-\dfrac{1}{2}\right)\)
Vậy phương trình cho vô nghiệm
Giải các hệ phương trình sau:
a) \(\left\{{}\begin{matrix}4x^2-4xy-14x-3y^2+y+10=0\\5\sqrt{xy}+2x+2y=6\sqrt{y}-8\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}2x^4+3x^2y+4x^2-2y^2+3y+2=0\\\sqrt{x\left(y-1\right)}+2y+2\sqrt{y-1}=3x+2\sqrt{x}+2\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}x^6+3x^2-y^3-6y^2-15y-14=0\\\sqrt{xy+2x-y-2}+6x-2y=10\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}xy+x+y=x^2-2y^2\\x\sqrt{2y}-y\sqrt{x-1}=2x-2y\end{matrix}\right.\)
Giải các hệ phương trình sau
\(1)\left\{{}\begin{matrix}\sqrt{x+1}=\sqrt{2}\left(8y^2+8y+1\right)\\4\left(x^3-8y^3\right)-6\left(x^2+4y^2\right)+3\left(x+2y\right)-1=0\end{matrix}\right.\)
\(2)\left\{{}\begin{matrix}3\sqrt{17x^2-y^2-6x+4}+x=6\sqrt{2x^2+x+y}-3y+2\\\sqrt{3x^2+xy+1}=\sqrt{x+1}\end{matrix}\right.\)
\(3)\left\{{}\begin{matrix}x^3+\left(2-y\right)x^2+\left(2-3y\right)x=5\left(x+1\right)\\3\sqrt{y+1}=3x^2-14x+14\end{matrix}\right.\)
\(4)\left\{{}\begin{matrix}4x^2=\left(\sqrt{x^2+1}+1\right)\left(x^2-y^3+3y-2\right)\\x^2+\left(y+1\right)^2=2\left(1+\dfrac{1-x^2}{y}\right)\end{matrix}\right.\)
\(5)\left\{{}\begin{matrix}7x^3+y^3+3xy\left(x-y\right)-12x^2+6x-1=0\\y^2+7y-17=9x+2\left(x+6\right)\sqrt{5-2y}\end{matrix}\right.\)
\(6)\left\{{}\begin{matrix}2x^2+3=4\left(x^2-2yx^2\right)\sqrt{3-2y}+\dfrac{4x^2+1}{x}\\\left(2x+1\right)\sqrt{2-\sqrt{3-2y}}=\sqrt[3]{2x^2+x^3}+x+2\end{matrix}\right.\)
giải hệ phương trình
a) \(\left\{{}\begin{matrix}\sqrt{2x^2+2y^2}+\sqrt{\frac{4}{3}\left(x^2+xy+y^2\right)}=2\left(x+y\right)\\\sqrt{3x+1}+\sqrt{5x+4}=3xy-y+3\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\\\sqrt{x+2y+1}+2\sqrt[3]{12x+7y+8}=2xy+x+5\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x^2+xy+x+3=0\\\left(x+1\right)^2+3\left(y+1\right)+2\left(xy-\sqrt{x^2y+2y}\right)=0\end{matrix}\right.\)
b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)
\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)
\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)
\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)
\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)
\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)
caau a) binh phuong len ra no x=y tuong tu
c)
ĐK $y \geqslant 0$
Hệ đã cho tương đương với
$\left\{\begin{matrix} 2x^2+2xy+2x+6=0\\ (x+1)^2+3(y+1)+2xy=2\sqrt{y(x^2+2)} \end{matrix}\right.$
Trừ từng vế $2$ phương trình ta được
$x^2+2+2\sqrt{y(x^2+2)}-3y=0$
$\Leftrightarrow (\sqrt{x^2+2}-\sqrt{y})(\sqrt{x^2+2}+3\sqrt{y})=0$
$\Leftrightarrow x^2+2=y$
Giải các hệ phương trình sau bằng phương pháp cộng đại số:
a) \(\left\{{}\begin{matrix}\sqrt{2}x-y=3\\x+\sqrt{2}y=\sqrt{2}\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{x}{2}-2y=\dfrac{3}{4}\\2x+\dfrac{y}{3}=-\dfrac{1}{3}\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\dfrac{2x-3y}{4}-\dfrac{x+y-1}{5}=2x-y-1\\\dfrac{x+y-1}{3}+\dfrac{4x-y-2}{4}=\dfrac{2x-y-3}{6}\end{matrix}\right.\)
a) Ta có: \(\left\{{}\begin{matrix}\sqrt{2}x-y=3\\x+\sqrt{2}y=\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2}x-y=3\\\sqrt{2}x+2y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-3y=1\\x+\sqrt{2}y=\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{3}\\x=\sqrt{2}-\sqrt{2}y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{3}\\x=\sqrt{2}-\sqrt{2}\cdot\dfrac{-1}{3}=\dfrac{4\sqrt{2}}{3}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{4\sqrt{2}}{3}\\y=-\dfrac{1}{3}\end{matrix}\right.\)
b) Ta có: \(\left\{{}\begin{matrix}\dfrac{x}{2}-2y=\dfrac{3}{4}\\2x+\dfrac{y}{3}=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-8y=3\\2x+\dfrac{1}{3}y=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{25}{3}y=\dfrac{10}{3}\\2x-8y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{2}{5}\\2x=3+8y=3+8\cdot\dfrac{-2}{5}=-\dfrac{1}{5}\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=-\dfrac{1}{10}\\y=-\dfrac{2}{5}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-\dfrac{1}{10}\\y=-\dfrac{2}{5}\end{matrix}\right.\)
c) Ta có: \(\left\{{}\begin{matrix}\dfrac{2x-3y}{4}-\dfrac{x+y-1}{5}=2x-y-1\\\dfrac{x+y-1}{3}+\dfrac{4x-y-2}{4}=\dfrac{2x-y-3}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5\left(2x-3y\right)}{20}-\dfrac{4\left(x+y-1\right)}{20}=\dfrac{20\left(2x-y-1\right)}{20}\\\dfrac{4\left(x+y-1\right)}{12}+\dfrac{3\left(4x-y-2\right)}{12}=\dfrac{2\left(2x-y-3\right)}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}10x-15y-4x-4y+4=40x-20y-20\\4x+4y-4+12x-3y-6=4x-2y-6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-19y+4-40x+20y+20=0\\16x+y-10-4x+2y+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-34x+y=-24\\12x+3y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-102x+3y=-72\\12x+3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-114x=-76\\12x+3y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\12\cdot\dfrac{2}{3}+3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\3y=4-8=-4\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{4}{3}\end{matrix}\right.\)
Giải hệ phương trình:
\(a,\left\{{}\begin{matrix}2x^3+x^2y+2x^2+xy+6=0\\x^2+3x+y=1\end{matrix}\right.\)
\(b,\left\{{}\begin{matrix}x^2=\left(2-y\right)\left(2+y\right)\\2x^3=\left(x+y\right)\left(4-xy\right)\end{matrix}\right.\)
\(c,\left\{{}\begin{matrix}\sqrt[3]{x+2y}=4-x-y\\\sqrt[3]{x+6}+\sqrt{2y}=2\end{matrix}\right.\)
a/
\(\Leftrightarrow\left\{{}\begin{matrix}x^2\left(2x+y\right)+x\left(2x+y\right)=-6\\x^2+x+2x+y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+x\right)\left(2x+y\right)=-6\\x^2+x+2x+y=1\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x^2+x=a\\2x+y=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}ab=-6\\a+b=1\end{matrix}\right.\) với
Theo Viet đảo, a và b là nghiệm của:
\(t^2-t-6=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+x=3\\2x+y=-2\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+x=-2\left(vn\right)\\2x+y=3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x-3=0\\y=-2x-2\end{matrix}\right.\) (bấm casio)
b/
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=4-y^2\\2x^3=\left(x+y\right)\left(4-xy\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=4\\2x^3=\left(x+y\right)\left(4-xy\right)\end{matrix}\right.\)
\(\Rightarrow2x^3=\left(x+y\right)\left(x^2+y^2-xy\right)\)
\(\Leftrightarrow2x^3=x^3+y^3\)
\(\Leftrightarrow x^3=y^3\Rightarrow x=y\)
Thay vào pt đầu:
\(2x^2=4\Rightarrow x^2=2\Rightarrow x=y=\pm\sqrt{2}\)
Giải hệ phương trình :
\(\left\{{}\begin{matrix}\sqrt{3+2x^2y-x^4y^2}+x^4\left(1-2x^2\right)=y^2\\1+\sqrt{1+\left(x-y\right)^2}=x^3\left(x^3-x+2y^2\right)\end{matrix}\right.\)
Gõ đề có sai không ạ?
\(\left\{{}\begin{matrix}\sqrt{3+2x^2y-x^4y^2}+x^4\left(1-2x^2\right)=y^4\\1+\sqrt{1+\left(x-y\right)^2}=x^3\left(x^3-x+2y^2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{4-\left(1-x^2y\right)^2}=2x^6-x^4+y^4\\-\sqrt{1+\left(x-y\right)^2}=1-x^6+x^4-2x^3y^2\end{matrix}\right.\)
Cộng theo vế HPT2
\(\sqrt{4-\left(1-x^2y\right)^2}-\sqrt{1+\left(x-y\right)^2}=\left(x^3-y^2\right)^2+1\)
\(\Leftrightarrow\sqrt{4-\left(1-x^2y\right)^2}=\sqrt{1+\left(x-y\right)^2}+\left(x^3-y^2\right)^2+1\) (1)
Có:
\(\left\{{}\begin{matrix}\sqrt{4-\left(1-x^2y\right)^2}\le2\\\sqrt{1+\left(x-y\right)^2}+\left(x^2-y^2\right)^2+1\ge2\end{matrix}\right.\)
\(\Rightarrow\) (1) xảy ra \(\Leftrightarrow\) \(\left\{{}\begin{matrix}\sqrt{4-\left(1-x^2y\right)^2}=2\\\sqrt{1+\left(x-y\right)^2}=1\\\left(x^3-y^2\right)^2=0\end{matrix}\right.\Leftrightarrow x=y=1\)