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Phương trình hoành độ giao điểm:
\(mx+2=x+m-1\)
\(\Leftrightarrow\left(m-1\right)x=m-3\)
\(d_1\) cắt \(d_2\) khi \(m\ne1\)
Khi đó: \(x=\dfrac{m-3}{m-1}=1-\dfrac{2}{m-1}\in Z\)
\(\Rightarrow m-1=Ư\left(2\right)=\left\{-2;-1;1;2\right\}\)
\(\Rightarrow m=\left\{-1;0;2;3\right\}\)
Do \(y=mx+2\Rightarrow\) nếu m, x đều nguyên thì y luôn nguyên (thỏa mãn)
Vậy \(m=\left\{-1;0;2;3\right\}\)
\(2,ĐK:\left\{{}\begin{matrix}\dfrac{2}{a+5}\ge0\\a+5\ne0\end{matrix}\right.\Leftrightarrow a+5>0\Leftrightarrow a>-5\left(C\right)\\ 3,M=2\sqrt{3}=\sqrt{12}< \sqrt{15}=N\left(C\right)\\ 4,=\left|3-\sqrt{3}\right|=3-\sqrt{3}\left(A\right)\\ 5,=\dfrac{3\sqrt{5}-3\sqrt{3}+3\sqrt{5}+3\sqrt{3}}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\dfrac{6\sqrt{5}}{2}=3\sqrt{5}\left(C\right)\)
Bài 5:
a, Áp dụng PTG: \(BC=\sqrt{AB^2+AC^2}=5\left(cm\right)\)
\(\sin B=\dfrac{AC}{BC}=\dfrac{3}{5}\approx\sin37^0\\ \Rightarrow\widehat{B}\approx37^0\\ \Rightarrow\widehat{C}\approx90^0-37^0=53^0\)
b, Áp dụng HTL: \(S_{AHC}=\dfrac{1}{2}AH\cdot HC=\dfrac{1}{2}\cdot\dfrac{AB\cdot AC}{BC}\cdot\dfrac{AC^2}{BC}=\dfrac{1}{2}\cdot\dfrac{12}{5}\cdot\dfrac{9}{5}=\dfrac{54}{25}\left(cm^2\right)\)
c, Vì AD là p/g nên \(\dfrac{DH}{DB}=\dfrac{AH}{AB}\)
Mà \(AC^2=CH\cdot BC\Leftrightarrow\dfrac{HC}{AC}=\dfrac{AC}{BC}\)
Mà \(AH\cdot BC=AB\cdot AC\Leftrightarrow\dfrac{AH}{AB}=\dfrac{AC}{BC}\)
Vậy \(\dfrac{DH}{DB}=\dfrac{HC}{AC}\)
7:
a: ĐKXĐ: x>=0; x<>1
\(D=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\)
\(=\dfrac{1}{2\left(\sqrt{x}-1\right)}-\dfrac{1}{2\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{-1}{\sqrt{x}+1}\)
b: Khi x=4/9 thì \(D=\dfrac{-1}{\dfrac{2}{3}+1}=-1:\dfrac{5}{3}=-\dfrac{3}{5}\)
c: |D|=1/3
=>D=-1/3 hoặc D=1/3
=>\(\left[{}\begin{matrix}\dfrac{-1}{\sqrt{x}+1}=\dfrac{-1}{3}\\\dfrac{-1}{\sqrt{x}+1}=\dfrac{1}{3}\left(loại\right)\end{matrix}\right.\)
=>\(\sqrt{x}+1=3\)
=>\(\sqrt{x}=2\)
=>x=4
6:
a: \(C=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)
\(=\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}-x+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(3-\sqrt{x}\right)}{2\sqrt{x}+4}\)
\(=\dfrac{3\left(\sqrt{x}+3\right)}{3+\sqrt{x}}\cdot\dfrac{-\sqrt{x}}{2\sqrt{x}+4}=\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)
b: C<-1
=>C+1<0
=>\(\dfrac{-3\sqrt{x}+2\sqrt{x}+4}{2\sqrt{x}+4}< 0\)
=>\(-\sqrt{x}+4< 0\)
=>\(-\sqrt{x}< -4\)
=>\(\sqrt{x}>4\)
=>x>16
\(C=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\\ =\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{1}{\sqrt{x}}\right)\\ =\left(\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}+\dfrac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\\ =\dfrac{3\sqrt{x}-x+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(3-\sqrt{x}\right)}{2\sqrt{x}+4}\\ =\dfrac{3\left(\sqrt{x}+3\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(3-\sqrt{x}\right)}{2\sqrt{x}+4}\\ =\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)
Để `C < -1` Ta có :
\(\dfrac{-3}{2\sqrt{x}+4}< -1\\ \Leftrightarrow\dfrac{-3}{2\sqrt{x}+4}+1< 0\\ \Leftrightarrow\dfrac{-3}{2\sqrt{x}+4}+\dfrac{2\sqrt{x}+4}{2\sqrt{x}+4}< 0\\ \Leftrightarrow-3+2\sqrt{x}+4< 0\\ \Leftrightarrow2\sqrt{x}+1< 0\\ \Leftrightarrow2\sqrt{x}< -1\\ \Leftrightarrow\sqrt{x}< -\dfrac{1}{2}\\ \Leftrightarrow x< \dfrac{1}{4}\)
b: Xét ΔABH vuông tại H có
\(AB^2=AH^2+HB^2\)
hay AH=12(cm)
Xét ΔAHB vuông tại H có
\(\sin\widehat{B}=\cos\widehat{C}=\dfrac{AH}{AB}=\dfrac{12}{13}\)
\(\cos\widehat{B}=\sin\widehat{C}=\dfrac{5}{13}\)
\(\tan\widehat{B}=\cot\widehat{C}=\dfrac{12}{5}\)
\(\cot\widehat{B}=\tan\widehat{C}=\dfrac{5}{12}\)
\(Q=x-2-2\sqrt{x-2}+4\)
\(=\left(\sqrt{x-2}-1\right)^2+3>=3\)
Dấu = xảy ra khi x=3
Câu 1:
a: Để hàm số đồng biến thì m-5>0
hay m>5
b: Để hàm số nghịch biến thì m-5<0
hay m<5
e: \(\dfrac{\left(5\sqrt{3}+5\sqrt{2}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)
\(=\dfrac{5\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}{5\left(\sqrt{3}-\sqrt{2}\right)}\)
\(=\sqrt{3}+\sqrt{2}\)