1.

\(cosA=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{1}{2}\Rightarrow\widehat{A}=60^o\)

\(S=\dfrac{1}{2}bc.sinA=\dfrac{1}{2}.8.5.sin60^o=10\sqrt{3}\)

\(S=\dfrac{1}{2}a.h_a=\dfrac{1}{2}.7.h_a=10\sqrt{3}\Rightarrow h_a=\dfrac{20\sqrt{3}}{7}\)

\(2R=\dfrac{a}{sinA}=\dfrac{7}{\dfrac{\sqrt{3}}{2}}=\dfrac{14\sqrt{3}}{3}\Rightarrow R=\dfrac{7\sqrt{3}}{3}\)

\(S=pr=\dfrac{a+b+c}{2}.r=10r=10\sqrt{3}\Rightarrow r=\sqrt{3}\)

\(m_a^2=\dfrac{b^2+c^2}{2}-\dfrac{a^2}{4}=\dfrac{129}{4}\Rightarrow m_a=\dfrac{\sqrt{129}}{2}\)

6.

a, Công thức trung tuyến:

\(AM^2=c^2=\dfrac{b^2+c^2}{2}-\dfrac{a^2}{4}=\dfrac{2b^2+2c^2-a^2}{4}\Rightarrow a^2=2\left(b^2-c^2\right)\)

b, \(a^2=2\left(b^2-c^2\right)\Rightarrow\dfrac{2\left(b^2-c^2\right)}{a^2}=1\)

\(\Leftrightarrow2\left(\dfrac{b^2}{a^2}-\dfrac{c^2}{a^2}\right)=1\)

\(\Leftrightarrow2\left(\dfrac{b^2}{a^2}.sin^2A-\dfrac{c^2}{a^2}.sin^2A\right)=sin^2A\)

\(\Leftrightarrow2\left(sin^2B-sin^2C\right)=sin^2A\)

Hay \(sin^2A=2\left(sin^2B-sin^2C\right)\)

Câu 7: Hình bạn tự vẽ

Giả sử hai trung tuyến \(AA_1;BB_1\) vuông góc với nhau. Gọi I là giao điểm của hai trung tuyến

Ta có: \(IA^2+IB^2=AB^2\)\(\Leftrightarrow\left(\dfrac{2}{3}m_a\right)^2+\left(\dfrac{2}{3}m_b\right)^2=c^2\)\(\Leftrightarrow4m_a^2+4m_b^2=9c^2\Leftrightarrow2\left(b^2+c^2\right)-a^2+2\left(a^2+c^2\right)-b^2=9c^2\Leftrightarrow a^2+b^2=5c^2\)(đúng với giả thiết)

suy ra giả sử là đúng . Vậy hai trung tuyến \(AA_1;BB_1\) vuông góc với nhau

  Câu 32 ạ

bài 2:

T₁= 27+273= 300 (K) 

bài 2:

T₁= 27+273= 300 (K) ; V₁=5m³

T₂=127+273=400 (K)  ; V₂=?

quá trình đẳng áp , áp suất p không đổi. Áp dụng mối liên hệ giữu thể tích và nhiệt độ tuyệt đối T ta có: 

\(\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\Rightarrow V_2=T_2\dfrac{V_1}{T_1}=400\dfrac{5}{300}\simeq6.67\left(m^3\right)\)

Okie, xinh nên giúp :3 Đùa thui

a/ 5 nguồn mắc nối tiếp \(\left\{{}\begin{matrix}\xi_b=5.\xi=5.4=20\left(V\right)\\r_b=5r=5.0,2=1\left(\Omega\right)\end{matrix}\right.\)

b/ \(R_D=\dfrac{U^2_{dm}}{P_{dm}}=\dfrac{36}{6}=6\left(\Omega\right);I_{dm}=\dfrac{P_{dm}}{U_{dm}}=\dfrac{6}{6}=1\left(A\right)\)

Đèn sáng bình thường \(\Rightarrow I_2=I_D=I_{dm}=1\left(A\right)\)

\(\left(R_1ntR_B\right)//\left(R_2ntR_D\right)\Rightarrow R_{td}=\dfrac{\left(R_1+R_B\right)\left(R_2+R_D\right)}{R_1+R_B+R_2+R_D}=\dfrac{\left(2+4\right)\left(6+6\right)}{2+4+6+6}=4\left(\Omega\right)\)

c/ \(I=\dfrac{\xi_b}{r_b+R_{td}}=\dfrac{20}{1+4}=4\left(A\right)\)

\(I=I_1+I_2\Rightarrow I_1=I-I_2=4-1=3\left(A\right)\Rightarrow P_1=I_1^2.R_1=3^2.2=18\left(W\right)\)

\(m_{Cu}=\dfrac{A_{Cu}.I_B.t}{F.n}=\dfrac{64.3.\left(32.60+10\right)}{96500.2}=...\left(g\right)\)

Đề bài là: Tính cos2x 

Cảm ơn mn nhiều ạ!

`sin3x sinx+sin(x-π/3) cos (x-π/6)=0`

`<=> 1/2 (cos2x - cos4x) + 1/2(-sin π/6 + sin (2x-π/2)=0`

`<=> cos2x-cos4x-1/2+ sin(2x-π/2)=0`

`<=>cos2x-cos4x-1/2+ sin2x .cos π/2 - cos2x. sinπ/2=0`

`<=> cos2x - cos4x - cos2x = 1/2`

`<=> cos4x = cos(2π)/3`

`<=>` \(\left[{}\begin{matrix}4x=\dfrac{2\text{π}}{3}+k2\text{π}\\4x=\dfrac{-2\text{π}}{3}+k2\text{π}\end{matrix}\right.\)

`<=>` \(\left[{}\begin{matrix}x=\dfrac{\text{π}}{6}+k\dfrac{\text{π}}{2}\\x=-\dfrac{\text{π}}{6}+k\dfrac{\text{π}}{2}\end{matrix}\right.\)

\(sin3x.sinx+sin\left(x-\dfrac{\pi}{3}\right)cos\left(x-\dfrac{\pi}{6}\right)=0\)

\(\Leftrightarrow\dfrac{1}{2}cos2x-\dfrac{1}{2}cos4x+\dfrac{1}{2}sin\left(2x-\dfrac{\pi}{2}\right)+\dfrac{1}{2}sin\left(-\dfrac{\pi}{6}\right)=0\)

\(\Leftrightarrow\dfrac{1}{2}cos2x-\dfrac{1}{2}cos4x-\dfrac{1}{2}cos2x-\dfrac{1}{4}=0\)

\(\Leftrightarrow cos4x+\dfrac{1}{2}=0\)

\(\Leftrightarrow2cos^22x-1+\dfrac{1}{2}=0\)

\(\Leftrightarrow cos^22x=\dfrac{1}{4}\)

\(\Rightarrow cos2x=\pm\dfrac{1}{2}\)

1) Vì x=25 thỏa mãn ĐKXĐ nên Thay x=25 vào biểu thức \(A=\dfrac{\sqrt{x}-2}{x+1}\), ta được:

\(A=\dfrac{\sqrt{25}-2}{25+1}=\dfrac{5-2}{25+1}=\dfrac{3}{26}\)

Vậy: Khi x=25 thì \(A=\dfrac{3}{26}\)

2) Ta có: \(B=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}+\dfrac{2x+8\sqrt{x}-6}{x-\sqrt{x}-2}\)

\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}+\dfrac{2x+8\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-5\sqrt{x}+6+2x+8\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

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

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

\(=\dfrac{3\sqrt{x}}{\sqrt{x}-2}\)

câu 3 chứ