HOC24
Lớp học
Môn học
Chủ đề / Chương
Bài học
1.
Q=\(\sqrt{4\left(1+6x+9x^2\right)}=\sqrt{2^2.\left[1^2+2.1.3x+\left(3x\right)^2\right]}=2\sqrt{\left(1+3x\right)^2}=2.\left|3x+1\right|\left(1\right)\)
Nếu \(x\ge-\frac{1}{3}\) thì (1)\(=2\left(3x+1\right)=6x+2\)
Nếu x<\(-\frac{1}{3}\) thì (1)=\(2\left(-3x-1\right)=-6x-2\)
2.
\(3^{3x-2}=81\Leftrightarrow3^{3x-2}=3^4\Leftrightarrow3x-2=4\Leftrightarrow3x=6\Leftrightarrow x=2\)
Vậy S={2}
ĐK: \(-3\le x\le6\)
\(\sqrt{x+3}+\sqrt{6-x}-\sqrt{\left(x-3\right)\left(6-x\right)}=3\)(1)
Đặt a=\(\sqrt{x+3}\left(a\ge0\right)\),b=\(\sqrt{6-x}\left(b\ge0\right)\)\(\Leftrightarrow a^2+b^2=9\)
Vậy (1)\(\Leftrightarrow a+b-ab=3\)
Vậy ta có hệ phương trình \(\left\{{}\begin{matrix}a^2+b^2=9\\a+b-ab=3\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left(a+b\right)^2-2ab=9\\a+b-ab=3\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left(a+b\right)^2-2ab=9\\2\left(a+b\right)-2ab=6\end{matrix}\right.\)\(\Leftrightarrow\left(a+b\right)^2+2\left(a+b\right)-15=0\Leftrightarrow\left(a+b-3\right)\left(a+b+5\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}a+b-3=0\\a+b+5=0\left(ktm\right)\end{matrix}\right.\)\(\Leftrightarrow a+b=3\)
Vậy \(\sqrt{x+3}+\sqrt{6-x}=3\)
Mà \(\sqrt{x+3}+\sqrt{6-x}-\sqrt{\left(x+3\right)\left(6-x\right)}=3\)
Suy ra \(\sqrt{\left(x+3\right)\left(6-x\right)}=0\Leftrightarrow\)\(\left[{}\begin{matrix}x+3=0\\6-x=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\x=6\end{matrix}\right.\)
Vậy S={-3;6}
\(\left\{{}\begin{matrix}x^2+y^2=2\left(xy+2\right)\\x+y=6\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x^2-2xy+y^2=4\\x+y=6\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left(x-y\right)^2=4\\x+y=6\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x-y=2\\x-y=-2\end{matrix}\right.\\x+y=6\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x-y=2\\x+y=6\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=4\\y=2\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x-y=-2\\x+y=6\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
Vậy (x;y)={(2;4);(4;2)}
\(x+\left|x-1\right|>5\)
TH1: x\(\ge1\)
Thì \(x+\left|x-1\right|>5\Leftrightarrow x+x-1>5\Leftrightarrow2x>6\Leftrightarrow x>3\)
TH2: x<1
Thì \(x+\left|x-1\right|>5\Leftrightarrow x+1-x>5\Leftrightarrow1>5\)(vô lí)
Vậy x>3 là nghiệm của phương trình
P=\(\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{x-4}\right):\left(\frac{\sqrt{x}-3}{2\sqrt{x}-x}\right)=\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}+\frac{4x}{4-x}\right).\frac{2\sqrt{x}-x}{\sqrt{x}-3}=\left[\frac{\left(2+\sqrt{x}\right)^2}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}-\frac{\left(2-\sqrt{x}\right)^2}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}+\frac{4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right].\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}=\frac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}.\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}=\frac{\left(4x+8\sqrt{x}\right).\sqrt{x}.\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)\left(\sqrt{x}-3\right)}=\frac{4x\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)\left(\sqrt{x}-3\right)}=\frac{4x}{\sqrt{x}-3}\)
ĐK:\(x>0,x\ne1\)
a) \(A=\left(\frac{1}{x-\sqrt{x}}+\frac{x+\sqrt{x}+1}{x\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}=\left[\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right]:\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}=\left[\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right].\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\left[\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right].\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}}\)
\(A=\frac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}=\frac{\sqrt{2}\left(3+\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{2}+\sqrt{3+\sqrt{5}}\right)}+\frac{\sqrt{2}\left(3-\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)}=\frac{\sqrt{2}\left(3+\sqrt{5}\right)}{2+\sqrt{6+2\sqrt{5}}}+\frac{\sqrt{2}\left(3-\sqrt{5}\right)}{2-\sqrt{6-2\sqrt{5}}}=\frac{\sqrt{2}\left(3+\sqrt{5}\right)}{2+\sqrt{\left(\sqrt{5}+1\right)^2}}+\frac{\sqrt{2}\left(3-\sqrt{5}\right)}{2-\sqrt{\left(\sqrt{5}-1\right)^2}}=\frac{\sqrt{2}\left(3+\sqrt{5}\right)}{2+\sqrt{5}+1}+\frac{\sqrt{2}\left(3-\sqrt{5}\right)}{2-\sqrt{5}+1}=\frac{\sqrt{2}\left(3+\sqrt{5}\right)}{3+\sqrt{5}}+\frac{\sqrt{2}\left(3-\sqrt{5}\right)}{3-\sqrt{5}}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
Gọi tam giác đã cho là △ABC vuông tại A (AB<AC) và BD và CE lần lượt là phân giác\(\Rightarrow\frac{AD}{DC}=\frac{3}{5},\frac{AE}{EB}=\frac{4}{5}\)(vì AB<AC)
\(\Rightarrow\frac{AB}{BC}=\frac{3}{5},\frac{AC}{BC}=\frac{4}{5}\)\(\Rightarrow AB:AC:BC=3:4:5\)
Mà AB+AC+BC=72
Suy ra \(\left\{{}\begin{matrix}AB=18\\AC=24\\BC=30\end{matrix}\right.\)
Vậy các cạnh của tam giác lần lượt là 18,24,30
Gọi x(km/h) là vận tốc của xe máy (x>0)
Vận tốc của xe ô tô là: x+15(km/h)
Quãng đường xe máy đã đi khi ô tô xuất phát là: \(90-\left[\left(6,5-6\right).x\right]=90-\frac{x}{2}=\frac{180-x}{2}\)(km)
Thời gian xe máy đến B là: \(\frac{180-x}{2x}\)(h)
Thời gian ô tô đến B là: \(\frac{90}{x+15}\)(h)
Vì cả hai xe đều đến B cùng lúc nên ta có phương trình \(\frac{180-x}{2x}=\frac{90}{x+15}\Leftrightarrow180x+2700-x^2-15x=180x\Leftrightarrow x^2+15x-2700=0\Leftrightarrow\left(x-45\right)\left(x+60\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=45\left(tm\right)\\x=-60\left(ktm\right)\end{matrix}\right.\)
Vậy vận tốc của xe máy là 45km/h
vận tốc của xe ô tô là 45+15=60km/h
\(\frac{a\sqrt{2}+2\sqrt{a}}{\sqrt{2}a}:\frac{1}{\sqrt{a}-\sqrt{2}}=\frac{\sqrt{2a}\left(\sqrt{a}+\sqrt{2}\right)}{\sqrt{2}a}.\left(\sqrt{a}-2\right)=\frac{\left(\sqrt{a}+\sqrt{2}\right)\left(\sqrt{a}-\sqrt{2}\right)}{\sqrt{a}}=\frac{a-2}{\sqrt{a}}\)