c.

Gọi E là trung điểm AD \(\Rightarrow EM\) là đường trung bình tam giác SAD

\(\Rightarrow\left\{{}\begin{matrix}EM=\dfrac{1}{2}SA=a\\EM||SA\Rightarrow EM\perp\left(ABCD\right)\end{matrix}\right.\)

\(\Rightarrow EC\) là hình chiếu vuông góc của CM lên (ABCD)

\(\Rightarrow\widehat{MCE}\) là góc giữa SM và (ABCD)

\(ED=\dfrac{1}{2}AD=a\Rightarrow EC=\sqrt{CD^2+ED^2}=a\sqrt{2}\)

\(\Rightarrow tan\widehat{MCE}=\dfrac{EM}{EC}=\dfrac{\sqrt{2}}{2}\Rightarrow\widehat{MCE}=...\)

e.

Gọi O là trung điểm BD, qua A kẻ đường thẳng song song BD cắt OE kéo dài tại F

\(\Rightarrow ABOF\) là hình bình hành (2 cặp cạnh đối song song)

\(\Rightarrow\left\{{}\begin{matrix}AF=OB=\dfrac{1}{2}BD\\AF||BD\end{matrix}\right.\)

Lại có MN là đường trung bình tam giác SBD \(\Rightarrow\left\{{}\begin{matrix}MN=\dfrac{1}{2}BD\\MN||BD\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}MN=AF\\MN||AF\end{matrix}\right.\) \(\Rightarrow ANMF\) là hình bình hành

\(\Rightarrow AN||MF\Rightarrow\left(AN;CM\right)=\left(AN;MF\right)=\widehat{CMF}\) nếu nó ko tù hoặc bằng góc bù của nó nếu \(\widehat{CMF}\) là góc tù

Ta có: \(MF=AN=\dfrac{a\sqrt{5}}{2}\) ; \(CM=\sqrt{CE^2+EM^2}=a\sqrt{3}\)

ABOF là hình bình hành nên AODF cũng là hình bình hành \(\Rightarrow E\) là tâm hình bình hành

\(\Rightarrow EF=OF=\dfrac{AB}{2}=\dfrac{a}{2}\)

Gọi G là giao điểm OE và BC \(\Rightarrow FG=EG+EF=a+\dfrac{a}{2}=\dfrac{3a}{2}\)

\(\Rightarrow CF=\sqrt{FG^2+CG^2}=\dfrac{a\sqrt{13}}{2}\)

ĐỊnh lý hàm cos:

\(cos\widehat{CMF}=\dfrac{CM^2+MF^2-CF^2}{2CM.MF}=\dfrac{\sqrt{15}}{15}\Rightarrow\widehat{CMF}\)

a) \(\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right)\)

\(=\left(2x+3\right)\left[\left(2x\right)^2-2x\cdot3+3^2\right]-2\left(4x^3-1\right)\)

\(=\left[\left(2x\right)^3+3^3\right]-2\left(4x^3-1\right)\)

\(=\left(8x^3+27\right)-8x^3+2\)

\(=8x^3+27-8x^3+2\)

\(=29\)

Vậy: ....

c) \(2\left(x^3+y^3\right)-3\left(x^3+y^3\right)\)

\(=2\left(x+y\right)\left(x^2-xy+y^2\right)-3x^2-3y^2\)

\(=2\left(x^2-xy+y^2\right)\cdot1-3x^2-3y^2\)

\(=2x^2-2xy+2y^2-3x^2-3y^2\)

\(=-x^2-2xy-y^2\)
\(=-\left(x^2+2xy+y^2\right)\)

\(=-\left(x+y\right)^2\)

\(=-\left(1\right)^2=-1\)

Vậy: ...

Sửa đề thành 96 cho dễ làm nha

\(\left(x-3\right)\left(x+1\right)\left(x+2\right)\left(x+6\right)=96\)

\(\Leftrightarrow\left[\left(x-3\right)\left(x+6\right)\right]\left[\left(x+1\right)\left(x+2\right)\right]=96\)

\(\Leftrightarrow\left(x^2+3x-18\right)\left(x^2+3x+2\right)=96\)

Đặt \(x^2-3x-8=a\)

<=> (a - 10) (a + 10) = 96

\(\Leftrightarrow a^2-100=96\)

\(\Leftrightarrow a^2=196\)

\(\Leftrightarrow\left[{}\begin{matrix}a=14\\a=-14\end{matrix}\right.\)

Giải típ đc chứ ??

a) \(A=\left(2\sqrt{12}-\sqrt{75}+\dfrac{1}{2}\sqrt{48}\right):\sqrt{3}\)

\(A=\left(4\sqrt{3}-5\sqrt{3}+2\sqrt{3}\right):\sqrt{3}\)

\(A=\sqrt{3}:\sqrt{3}\)

\(A=1\)

b) \(B=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(B=\left|2-\sqrt{5}\right|-\left|\sqrt{5}+1\right|\)

\(B=-2+\sqrt{5}-\sqrt{5}-1\)

\(B=-3\)

c) \(C=\dfrac{3}{\sqrt{7}-2}-\dfrac{4}{3+\sqrt{7}}\)

\(C=\dfrac{3\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\dfrac{4\left(3-\sqrt{7}\right)}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}\)

\(C=\dfrac{3\left(\sqrt{7}+2\right)}{3}-\dfrac{4\left(3-\sqrt{7}\right)}{2}\)

\(C=\sqrt{7}+2-2\left(3-\sqrt{7}\right)\)

\(C=\sqrt{7}+2-6+2\sqrt{7}\)

\(C=3\sqrt{7}-4\)

d) \(D=3\sqrt{2a}-\sqrt{18a^3}+4\sqrt{\dfrac{a}{2}}-\dfrac{1}{4}\sqrt{128a}\)

\(D=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-\dfrac{1}{4}\cdot8\sqrt{2a}\)

\(D=5\sqrt{2a}-3a\sqrt{2a}-2\sqrt{2a}\)

\(D=3\sqrt{2a}-3a\sqrt{2a}\)

e) \(E=\dfrac{3+\sqrt{3}}{\sqrt{3}}-\dfrac{2}{\sqrt{3}-1}\)

\(E=\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}}-\dfrac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(E=\left(\sqrt{3}+1\right)-\dfrac{2\left(\sqrt{3}+1\right)}{2}\)

\(E=\left(\sqrt{3}+1\right)-\left(\sqrt{3}+1\right)\)

\(E=0\)

Lời giải:

a. 

\(A=2\sqrt{\frac{12}{3}}-\sqrt{\frac{75}{3}}+\frac{1}{2}\sqrt{\frac{48}{3}}=2\sqrt{4}-\sqrt{25}+\frac{1}{2}\sqrt{16}\)

\(2.2-5+\frac{1}{2}.4=1\)

b. 

\(B=|2-\sqrt{5}|-|\sqrt{5}+1|=\sqrt{5}-2-(\sqrt{5}+1)=-3\)

c. 

\(C=\frac{3(\sqrt{7}+2)}{(\sqrt{7}-2)(\sqrt{7}+2)}-\frac{4(3-\sqrt{7})}{(3+\sqrt{7})(3-\sqrt{7})}\)

\(=\frac{3(\sqrt{7}+2)}{7-2^2}-\frac{4(3-\sqrt{7})}{3^2-7}\)

\(=\frac{3(\sqrt{7}+2)}{3}-\frac{4(3-\sqrt{7})}{2}=\sqrt{7}+2-2(3-\sqrt{7})=-4+3\sqrt{7}\)

e. 

\(E=\frac{\sqrt{3}(\sqrt{3}+1)}{\sqrt{3}}-\frac{2(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\sqrt{3}+1-\frac{2(\sqrt{3}+1)}{3-1^2}=(\sqrt{3}+1)-(\sqrt{3}+1)=0\)

Những câu đã đăng rồi thì em hạn chế đăng lại nhé.

a: \(=\dfrac{\left(4\sqrt{3}-5\sqrt{3}+\dfrac{1}{2}\cdot4\sqrt{3}\right)}{\sqrt{3}}\)

=4-5+1/2*4

=-1+2

=1

b: \(=\left|2-\sqrt{5}\right|-\left|\sqrt{5}+1\right|\)

\(=\sqrt{5}-2-\sqrt{5}-1=-3\)

c: \(=\dfrac{3\left(\sqrt{7}+2\right)}{3}-\dfrac{4\left(3-\sqrt{7}\right)}{2}\)

\(=\sqrt{7}+2-2\left(3-\sqrt{7}\right)\)

\(=\sqrt{7}+2-6+2\sqrt{7}=3\sqrt{7}-4\)

d: \(=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-\dfrac{1}{4}\cdot8\sqrt{2a}\)

\(=3\sqrt{2a}-3a\cdot\sqrt{2a}+2\sqrt{2a}-2\sqrt{2a}\)

\(=\sqrt{2a}\left(3-3a\right)\)

e: \(=\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}}-\dfrac{2\left(\sqrt{3}+1\right)}{2}\)

\(=\sqrt{3}-\sqrt{3}-1=-1\)

a, Ta có : \(\sin^2x+\cos^2x=1\)

\(\Rightarrow\sin x=\sqrt{1-\cos^2x}=\left|\dfrac{\sqrt{15}}{4}\right|\)

Mà \(0< x< \dfrac{\pi}{2}\)

\(\Rightarrow\sin x=\dfrac{\sqrt{15}}{4}\)

Ta lại có : \(\left\{{}\begin{matrix}\sin2x=2\sin x\cos x=\dfrac{\sqrt{15}}{8}\\\cos2x=2\cos^2x-1=-\dfrac{7}{8}\end{matrix}\right.\)

Vậy ...

c, Ta có : \(\tan2x=\dfrac{2\tan x}{1-\tan^2x}=\dfrac{4}{3}=\dfrac{\sin2x}{\cos2x}\)

- Ta có HPT : \(\left\{{}\begin{matrix}\sin^22x+\cos^22x=1\\3\sin2x-4\cos2x=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\sin2x=\left|\dfrac{4}{5}\right|\\\cos2x=\left|\dfrac{3}{5}\right|\end{matrix}\right.\)

Lại có : \(\pi< x< \dfrac{3}{2}\pi\)

\(\Rightarrow\left\{{}\begin{matrix}\sin2x=\dfrac{4}{5}\\\cos2x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy ...

\(a,m=0\Leftrightarrow y=3x+2\)

Vì \(3>0\) nên hàm đồng biến

\(b,\text{Thay }x=-1;y=3\\ \Leftrightarrow-m-3+2=3\Leftrightarrow m=-4\\ c,\text{PT giao Ox: }y=0\Leftrightarrow x=-\dfrac{2}{m+3}\Leftrightarrow A\left(-\dfrac{2}{m+3};0\right)\Leftrightarrow OA=\dfrac{2}{\left|m+3\right|}\\ \text{PT giao Oy: }x=0\Leftrightarrow y=2\Leftrightarrow B\left(0;2\right)\Leftrightarrow OB=2\\ \text{Ta có }S_{OAB}=4\\ \Leftrightarrow\dfrac{1}{2}OA\cdot OB=4\Leftrightarrow\dfrac{2}{\left|m+3\right|}\cdot2=8\\ \Leftrightarrow\dfrac{4}{\left|m+3\right|}=8\\ \Leftrightarrow\left|m+3\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}m=-\dfrac{5}{2}\\m=-\dfrac{7}{2}\end{matrix}\right.\)

1: Thay x=9 vào A, ta được:

\(A=\dfrac{\sqrt{9}-1}{\sqrt{9}}=\dfrac{3-1}{3}=\dfrac{2}{3}\)

2: \(B=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

3: P=B:A

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}:\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\)

Để P lớn nhất thì \(\dfrac{2}{\sqrt{x}-1}+1\) lớn nhất

=>\(\dfrac{2}{\sqrt{x}-1}\) lớn nhất

=>\(\sqrt{x}-1\) là số nguyên dương nhỏ nhất

=>\(\sqrt{x}-1=1\)

=>\(\sqrt{x}=2\)

=>x=4(nhận)

\(b,\Leftrightarrow\left\{{}\begin{matrix}a=2;b\ne3\\2a+b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=-3\end{matrix}\right.\\ b,\text{Gọi }M\left(x_0;y_0\right)\text{ là điểm cần tìm }\Leftrightarrow y_0=3x_0\\ M\left(x_0;y_0\right)\in\left(d\right)\Leftrightarrow2x_0+3=y_0=3x_0\Leftrightarrow x_0=3\Leftrightarrow y_0=9\\ \text{Vậy }M\left(3;9\right)\text{ là điểm cần tìm}\)