giai pt
\(9\left(\sqrt{x^3+8}-2\right)=2\left(x^2-1\right)\)
1. Cho pt: x2 -2(m+1)x+m2=0 (1). Tìm m để pt có 2 nghiệm x1 ; x2 thỏa mãn (x1-m)2 + x2=m+2.
2. Giai pt: \(\left(x-1\right)\sqrt{2\left(x^2+4\right)}=x^2-x-2\)
3. Giai hệ pt: \(\left\{{}\begin{matrix}\frac{1}{\sqrt[]{x}}-\frac{\sqrt{x}}{y}=x^2+xy-2y^2\left(1\right)\\\left(\sqrt{x+3}-\sqrt{y}\right)\left(1+\sqrt{x^2+3x}\right)=3\left(2\right)\end{matrix}\right.\)
4. Giai pt trên tập số nguyên \(x^{2015}=\sqrt{y\left(y+1\right)\left(y+2\right)\left(y+3\right)}+1\)
Câu 1: Giai pt:
\(\sqrt{x+1}+\sqrt[3]{x^2+1}=2\)
Câu 2 : Gỉa hệ pt :
\(\left\{{}\begin{matrix}xy\left(x+y\right)-x\left(y-1\right)-\frac{3}{8}\left(y^2+1\right)=0\\xy^2+x-y=0\end{matrix}\right.\)
câu 1 ta dùng liên hợp nha bạn
điều kiện \(x\ge-1\)
\(\sqrt{x+1}-1+\sqrt[3]{x^2+1}-1=0\\ \Leftrightarrow\frac{x}{\sqrt{x+1}+1}+\frac{x^2}{\sqrt[3]{x+1}^2+\sqrt[3]{x+1}+1}=0\)
suy ra là \(\left[{}\begin{matrix}x=0\left(n\right)\\\frac{1}{\sqrt{x+1}+1}+\frac{x}{\sqrt[3]{x+1}^2+\sqrt[3]{x+1}+1}=0\left(1\right)\end{matrix}\right.\)
theo mình nghĩ (1) vô nghiệm
vậy x=0 là nghiệm pt
giải pt :a,\(\left(2x+6\right)\sqrt{x+4}-\left(x-5\right)\sqrt{2x+3}=3\left(x-1\right)\)
b, \(\left(4x+1\right)\sqrt{x+2}-\left(4x-1\right)\sqrt{x-2}=21\)
c, \(\left(4x+2\right)\sqrt{x+1}-\left(4x-2\right)\sqrt{x-1}=9\)
d, \(\left(2x-4\right)\sqrt{3x-2}+\sqrt{x+3}=5x-7+\sqrt{3x^2+7x-6}\)
giải pt :
a, \(\left(2x-6\right)\sqrt{x+4}-\left(x-5\right)\sqrt{2x+3}=3\left(x-1\right)\)
b, \(\left(4x+1\right)\sqrt{x+2}-\left(4x-1\right)\sqrt{x-2}=21\)
c, \(\left(4x+2\right)\sqrt{x+1}-\left(4x-2\right)\sqrt{x-1}=9\)
d, \(\left(2x-4\right)\sqrt{3x-2}+\sqrt{x+3}=5x-7+\sqrt{3x^2+7x-6}\)
Giai pt
\(\sqrt{2x+1}-2\sqrt{2-x}=3\sqrt[4]{\left(1-2x\right)\left(x-2\right)}\)
giai pt :
\(\sqrt{1+x}+\sqrt{8-x}+\sqrt{\left(1+x\right)\left(8-x\right)}=3\)
\(ĐK:-1\le x\le8\)
Đặt \(\sqrt{1+x}=u;\sqrt{8-x}=v\)thì \(\left(u+v\right)^2=9+2\sqrt{uv}\Rightarrow\sqrt{uv}=\frac{\left(u+v\right)^2-9}{2}\)
Phương trình lúc này có dạng \(\left(u+v\right)+\frac{\left(u+v\right)^2-9}{2}=3\Leftrightarrow\left(u+v\right)^2+2\left(u+v\right)-15=0\)\(\Leftrightarrow\left(u+v+5\right)\left(u+v-3\right)=0\Leftrightarrow\orbr{\begin{cases}u+v=-5\left(L\right)\\u+v=3\left(tm\right)\end{cases}}\)
Như vậy, \(u+v=3\Rightarrow\sqrt{uv}=\frac{3^2-9}{2}=0\Rightarrow uv=0\)
u, v là hai nghiệm của phương trình \(t^2-3t=0\Leftrightarrow\orbr{\begin{cases}t=3\\t=0\end{cases}}\)
* Nếu u = 3, v = 0 thì \(\hept{\begin{cases}\sqrt{1+x}=3\\\sqrt{8-x}=0\end{cases}}\Rightarrow x=8\left(tm\right)\)
* Nếu u = 0, v = 3 thì \(\hept{\begin{cases}\sqrt{1+x}=0\\\sqrt{8-x}=3\end{cases}}\Rightarrow x=-1\left(tm\right)\)
Vậy phương trình có tập nghiệm \(S=\left\{-1;8\right\}\)
thể giải thích chỗ \(\left(u+v\right)^2=9+2\sqrt{uv}\) đc ko
Giai he pt: \(\left\{{}\begin{matrix}\left(x-y\right)^2+4=3y-5x+2\sqrt{\left(x+1\right)\left(y-1\right)}\\\frac{3xy-5y-6x+11}{\sqrt{x^3+1}}=5\end{matrix}\right.\)
1\(\left\{{}\begin{matrix}xy\left(x+y\right)=2\\x^3+y^3+x^3y^3+7\left(x+1\right)\left(y+1\right)=31\end{matrix}\right.\)
2 giải pt \(9+3\sqrt{x\left(3-2x\right)}=7\sqrt{x}+5\sqrt{3-2x}\)
\(\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3-3xy\left(x+y\right)+\left(xy\right)^3+7\left(xy+x+y+1\right)=31\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3+\left(xy\right)^3+7\left(xy+x+y\right)=30\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\) với \(u^2\ge4v\)
\(\Rightarrow\left\{{}\begin{matrix}uv=2\\u^3+v^3+7\left(u+v\right)=30\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3-3uv\left(u+v\right)+7\left(u+v\right)=30\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3+\left(u+v\right)-30=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\u+v=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=2\\v=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=2\\xy=1\end{matrix}\right.\) \(\Leftrightarrow\left(x;y\right)=\left(1;1\right)\)
2.
ĐKXĐ: \(0\le x\le\dfrac{3}{2}\)
\(\Leftrightarrow9x\left(3-2x\right)+81+54\sqrt{x\left(3-2x\right)}=49x+25\left(3-2x\right)+70\sqrt{x\left(3-2x\right)}\)
\(\Leftrightarrow9x^2-14x-3+8\sqrt{x\left(3-2x\right)}=0\)
\(\Leftrightarrow9\left(x^2-2x+1\right)-4\left(3-x-2\sqrt{x\left(3-2x\right)}\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2-\dfrac{36\left(x-1\right)^2}{3-x+2\sqrt{x\left(3-2x\right)}}=0\)
\(\Leftrightarrow9\left(x-1\right)^2\left(1-\dfrac{4}{3-x+2\sqrt{x\left(3-2x\right)}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\3-x+2\sqrt{x\left(3-2x\right)}=4\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2\sqrt{x\left(3-2x\right)}=x+1\)
\(\Leftrightarrow4x\left(3-2x\right)=x^2+2x+1\)
\(\Leftrightarrow9x^2-10x+1=0\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)
Giai pt
\(\left(\sqrt{2x+3}+2\right).\left(\sqrt{x+6}-\sqrt{x+1}\right)=5\)
\(\left(\sqrt{2x+3}+2\right)\left(\sqrt{x+6}-\sqrt{x+1}\right)=5\)
\(ĐKXĐ:x\ge-1\).Nhận thấy \(\sqrt{x+6}-\sqrt{x+1}>0\)
\(\Leftrightarrow\left(\sqrt{2x+3}+2\right)\frac{\left(\sqrt{x+6}+\sqrt{x+1}\right)\left(\sqrt{x+6}-\sqrt{x+1}\right)}{\sqrt{x+6}-\sqrt{x+1}}=5\)
\(\Leftrightarrow\left(\sqrt{2x+3}+2\right)\frac{5}{\sqrt{x+6}-\sqrt{x+1}}=5\)
\(\Leftrightarrow\frac{\sqrt{2x+3}+2}{\sqrt{x+6}-\sqrt{x+1}}=1\)
\(\Leftrightarrow\sqrt{2x+3}+2-\sqrt{x+6}+\sqrt{x+1}=0\)
Th1:\(\sqrt{x+1}=2\Leftrightarrow x=3\left(thoaman\right)\)
Th2:\(\sqrt{x+1}-2\ne0\Leftrightarrow x\ne3\)
\(\Leftrightarrow\left(\sqrt{2x+3}-\sqrt{x+6}\right)+\left(2+\sqrt{x+1}\right)=0\)
\(\Leftrightarrow\frac{x-3}{\sqrt{2x+3}+\sqrt{x+6}}+\frac{x-3}{\sqrt{x+1}-2}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt{2x+3}+\sqrt{x+6}}+\frac{1}{\sqrt{x+1}-2}\right)=0\)
Tự lm tiếp nha