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Lời giải:
ĐK: $x\geq 0$
Đặt $\sqrt{x+1}=a; \sqrt{x}=b$. ĐK $a,b\geq 0$ thì ta có:
$a-b-ab=a^2-2b^2$
$\Leftrightarrow a-b=a^2+ab-2b^2=(a-b)(a+2b)$
$\Leftrightarrow (a-b)(a+2b-1)=0$
$\Leftrightarrow a=b$ hoặc $a+2b=1$
Nếu $a=b\Rightarrow a^2=b^2\Leftrightarrow x+1=x$ (vô lý)
Nếu $a+2b=1$
$\Leftrightarrow \sqrt{x+1}-1+2\sqrt{x}=0$
$\Leftrightarrow \frac{x}{\sqrt{x+1}+1}+2\sqrt{x}=0$
$\Leftrightarrow \sqrt{x}(\frac{\sqrt{x}}{\sqrt{x+1}+1}+2)=0$
Dễ thấy biểu thức trong ngoặc lớn hơn $0$ nên \sqrt{x}=0$
$\Leftrightarrow x=0$
Vậy.......
1) \(\Leftrightarrow4-4\sqrt{\dfrac{x+2}{x-3}}=x+7\)
\(\Leftrightarrow-4\sqrt{\dfrac{x+2}{x-3}}=x+3\)
\(\Leftrightarrow16\dfrac{x+2}{x-3}=x^2+6x+9\)
\(\Leftrightarrow16x+3=x^3+6x^2+9x-3x^2-18x-27\)
\(\Leftrightarrow x^3+3x^2-25x-59=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4,79\\x=-2,2\\x=-5,58\end{matrix}\right.\)
Vậy tập nghiệm....
ĐKXĐ: \(x\ge-1\)
\(x^2-1+\sqrt{x+1}=0\Rightarrow\left(x-1\right)\left(x+1\right)+\sqrt{x+1}=0\)
\(\Rightarrow\left(x+1-2\right)\left(x+1\right)+\sqrt{x+1}=0\)
Đặt \(\sqrt{x+1}=t\ge0\Rightarrow x+1=t^2\) ta được:
\(\left(t^2-2\right)t^2+t=0\Rightarrow t\left(\left(t^2-2\right)t+1\right)=0\)
\(\Rightarrow t\left(t^3-2t+1\right)=0\Rightarrow t\left(t-1\right)\left(t^2+t-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t=0\\t-1=0\\t^2+t-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}t=0\\t=1\\t=\dfrac{-1+\sqrt{5}}{2}\\t=\dfrac{-1-\sqrt{5}}{2}< 0\left(l\right)\end{matrix}\right.\)
TH1: \(t=0\Rightarrow\sqrt{x+1}=0\Rightarrow x=-1\)
TH2: \(t=1\Rightarrow\sqrt{x+1}=1\Rightarrow x+1=1\Rightarrow x=0\)
TH3: \(t=\dfrac{-1+\sqrt{5}}{2}\Rightarrow\sqrt{x+1}=\dfrac{-1+\sqrt{5}}{2}\Rightarrow x+1=\dfrac{3-\sqrt{5}}{2}\)
\(\Rightarrow x=\dfrac{3-\sqrt{5}}{2}-1=\dfrac{1-\sqrt{5}}{2}\)
Vậy pt có 3 nghiệm \(\left[{}\begin{matrix}x=-1\\x=0\\x=\dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\)
Lời giải:
Đặt \(\sqrt{x+1}=a\Rightarrow 1=a^2-x\)
PT trở thành: \(x^2+a=a^2-x\)
\(\Leftrightarrow x^2-a^2+(a+x)=0\)
\(\Leftrightarrow (x+a)(x-a+1)=0\Rightarrow \left[\begin{matrix} x=-a\\ x+1=a\end{matrix}\right.\)
Nếu \(x=-a=-\sqrt{x+1}\Rightarrow \left\{\begin{matrix} x\leq 0\\ x^2=x+1\end{matrix}\right.\Rightarrow x=\frac{1+\sqrt{5}}{2}\)
Nếu \(x+1=a=\sqrt{x+1}\Rightarrow (x+1)^2=(x+1)\Rightarrow x(x+1)=0\)
\(\Rightarrow \left[\begin{matrix} x=0\\ x=-1\end{matrix}\right.\) (đều thỏa mãn)
Vậy.........
a) \(3\sqrt{x^2+3x}=\left(x+5\right)\left(2-x\right)\)
\(\Leftrightarrow3\sqrt{x^2+3x}=-x^2-3x+10\)
\(\Leftrightarrow\left(x^2+3x\right)+3\sqrt{x^2+3x}-10=0\)
Đặt \(t=\sqrt{x^2+3x}\left(t\ge0\right)\left(1\right)\)
Ta có:
\(\Rightarrow t^2+3t-10=0\)
\(\Rightarrow t_1=2\left(TM\right);t_2=-5\left(KTM\right)\)
thay \(t=2\) vào (1), ta có :
\(\sqrt{x^2+3x}=2\)
\(\Leftrightarrow x^2+3x=4\Leftrightarrow x^2+3x-4=0\)
\(\Rightarrow x_1=1;x_2=-4\)
vậy phương trình có 3 nghiệm x1 = 1, x2 = -4
b) \(\sqrt{5x^2+10x+1}=7-x^2-2x\)
\(\Leftrightarrow\sqrt{5x^2+10x+1}=\left(5x^2+10x+1\right)-6x^2+12x-6\)
\(\Leftrightarrow\sqrt{5x^2+10x+1}=\left(5x^2+10x+1\right)-6\left(x-1\right)^2\)
Đặt \(t=\sqrt{5x^2+10x+1}\) (t lớn hơn hoặc bằng 0) (1)
ta có :...............
mk chỉ bt làm đến đấy thôi, hình như đây là ôn hsg toán 10 à
b/ \(\sqrt{12-\dfrac{12}{x^2}}+\sqrt{x^2-\dfrac{12}{x^2}}=x^2\)
\(\Leftrightarrow x-\sqrt{12-\dfrac{12}{x^2}}=\sqrt{x^2-\dfrac{12}{x^2}}\)
Bình phương 2 vế rút gọn
\(\Leftrightarrow x^4-x^2-4\sqrt{3\left(x^4-x^2\right)}+12=0\)
Đặt \(\sqrt{x^4-x^2}=a\)
\(\Rightarrow a^2-4\sqrt{3}a+12=0\)
\(\Leftrightarrow a=2\sqrt{3}\)
\(\Leftrightarrow x^4-x^2=12\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
1.
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y+x^3y+xy^2+xy=-\dfrac{5}{4}\\x^4+y^2+xy\left(1+2x\right)=-\dfrac{5}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+y\right)+xy+xy\left(x^2+y\right)=-\dfrac{5}{4}\\\left(x^2+y\right)^2+xy=-\dfrac{5}{4}\end{matrix}\right.\left(1\right)\)
Đặt \(\left\{{}\begin{matrix}x^2+y=a\\xy=b\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}a+b+ab=-\dfrac{5}{4}\\a^2+b=-\dfrac{5}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-a^2-\dfrac{5}{4}-a\left(a^2+\dfrac{5}{4}\right)=-\dfrac{5}{4}\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2-a^3-\dfrac{1}{4}a=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-a\left(a^2-a+\dfrac{1}{4}\right)=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a\left(a-\dfrac{1}{2}\right)^2=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=0\\b=-\dfrac{5}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=-\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}a=0\\b=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+y=0\\xy=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{\sqrt[3]{10}}{2}\\y=-\dfrac{5}{2\sqrt[3]{10}}\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+y=\dfrac{1}{2}\\xy=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-\dfrac{3}{2}\end{matrix}\right.\)
Kết luận: Phương trình đã cho có nghiệm \(\left(x;y\right)\in\left\{\left(\dfrac{\sqrt[3]{10}}{2};-\dfrac{5}{2\sqrt[3]{10}}\right);\left(1;-\dfrac{3}{2}\right)\right\}\)
2.
\(\left\{{}\begin{matrix}\left(x+1\right)^3-16\left(x+1\right)=\left(\dfrac{2}{y}\right)^3-4\left(\dfrac{2}{y}\right)\\1+\left(\dfrac{2}{y}\right)^2=5\left(x+1\right)^2+5\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+1=u\\\dfrac{2}{y}=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u^3-16u=v^3-4v\\v^2=5u^2+4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}u^3-v^3=16u-4v\\4=v^2-5u^2\end{matrix}\right.\)
\(\Rightarrow4\left(u^3-v^3\right)=\left(16u-4v\right)\left(v^2-5u^2\right)\)
\(\Leftrightarrow21u^3-5u^2v-4uv^2=0\)
\(\Leftrightarrow u\left(7u-4v\right)\left(3u+v\right)=0\Rightarrow\left[{}\begin{matrix}u=0\Rightarrow v^2=4\\u=\dfrac{4v}{7}\Rightarrow4=v^2-5\left(\dfrac{4v}{7}\right)^2\\v=-3u\Rightarrow4=\left(-3u\right)^2-5u^2\end{matrix}\right.\)
\(\Rightarrow...\)
Lời giải:
HPT tương đương:
\(\left\{\begin{matrix} 2x^2y=y^2+1\\ 2xy^2=x^2+1\end{matrix}\right.\)
Trừ hai pt cho nhau thì:
$2xy(x-y)+x^2-y^2=0$
$\Leftrightarrow 2xy(x-y)+(x-y)(x+y)=0$
$\Leftrightarrow (x-y)(2xy+x+y)=0$
$\Leftrightarrow x-y=0$ hoặc $2xy+x+y=0$
Nếu $x-y=0\Leftrightarrow x=y$. Thay vào pt (1):
$2x^2=x+\frac{1}{x}$
$\Rightarrow 2x^3=x^2+1$
$\Leftrightarrow (x-1)(2x^2+x+1)=0$
Đến đấy thì đơn giản rồi.
Nếu $2xy+x+y=0$:
Từ $2x^2=y+\frac{1}{y}=\frac{y^2+1}{y}$
Mà $2x^2>0; y^2+1>0$ với mọi $x,y\neq 0$ nên $y>0$
Tương tự $x>0$
$\Rightarrow 2xy+x+y>0$. Do đó TH này loại
Vậy...........
https://diendantoanhoc.net/topic/163051-x-fracxsqrtx2-1-frac3512/