\(A=\frac{tana+tanb}{tan\left(a+b\right)}-\frac{tana-tanb}{tan\left(a-b\right)}=\frac{tana+tanb}{\frac{tana+tanb}{1-tana\cdot tanb}}-\frac{tana-tanb}{\frac{tana-tanb}{1+tana\cdot tanb}}\\ \Leftrightarrow A=1-tana\cdot tanb-1-tana\cdot tanb=-2tana\cdot tanb\)

\(B=\frac{cos^3x-cos3x}{cosx}+\frac{sin^3x+sin3x}{sinx}\\ B=\frac{cos^3x-4cos^3x+3cosx}{cosx}+\frac{sin^3x+3sinx-4sin^3x}{sinx}\\ B=\frac{-3cos^3x+3cosx}{cosx}+\frac{-3sin^3x+3sinx}{sinx}\\ B=\frac{cosx\left(-3cos^2x+3\right)}{cosx}+\frac{sinx\left(-3sin^2x+3\right)}{sinx}\\ B=-3cos^2x+3-3sin^2x+3=6-3\left(sin^2x+cos^2x\right)=6-3=3\)

~~~~~~~~chúc bạn học tốt~~~~~~~~~~

Câu a)

Ta sử dụng 2 công thức:

\(\bullet \tan (180-\alpha)=-\tan \alpha\)

\(\bullet \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha.\tan \beta}\)

Áp dụng vào bài toán:

\(\text{VT}=\tan A+\tan B+\tan C=\tan A+\tan B+\tan (180-A-B)\)

\(=\tan A+\tan B-\tan (A+B)=\tan A+\tan B-\frac{\tan A+\tan B}{1-\tan A.\tan B}\)

\(=(\tan A+\tan B)\left(1+\frac{1}{1-\tan A.\tan B}\right)=(\tan A+\tan B).\frac{-\tan A.\tan B}{1-\tan A.\tan B}\)

\(=-\tan A.\tan B.\frac{\tan A+\tan B}{1-\tan A.\tan B}=-\tan A.\tan B.\tan (A+B)\)

\(=\tan A.\tan B.\tan (180-A-B)\)

\(=\tan A.\tan B.\tan C=\text{VP}\)

Do đó ta có đpcm

Tam giác $ABC$ có ba góc nhọn nên \(\tan A, \tan B, \tan C>0\)

Áp dụng BĐT Cauchy ta có:

\(P=\tan A+\tan B+\tan C\geq 3\sqrt[3]{\tan A.\tan B.\tan C}\)

\(\Leftrightarrow P=\tan A+\tan B+\tan C\geq 3\sqrt[3]{\tan A+\tan B+\tan C}\)

\(\Rightarrow P\geq 3\sqrt[3]{P}\)

\(\Rightarrow P^3\geq 27P\Leftrightarrow P(P^2-27)\geq 0\)

\(\Rightarrow P^2-27\geq 0\Rightarrow P\geq 3\sqrt{3}\)

Vậy \(P_{\min}=3\sqrt{3}\). Dấu bằng xảy ra khi \(\angle A=\angle B=\angle C=60^0\)

Câu b)

Ta sử dụng 2 công thức chính:

\(\bullet \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha.\tan \beta}\)

\(\bullet \tan (90-\alpha)=\frac{1}{\tan \alpha}\)

Áp dụng vào bài toán:

\(\text{VT}=\tan \frac{A}{2}.\tan \frac{B}{2}+\tan \frac{B}{2}.\tan \frac{C}{2}+\tan \frac{C}{2}.\tan \frac{A}{2}\)

\(=\tan \frac{A}{2}.\tan \frac{B}{2}+\tan \frac{C}{2}(\tan \frac{A}{2}+\tan \frac{B}{2})\)

\(=\tan \frac{A}{2}.\tan \frac{B}{2}+\tan (90-\frac{A+B}{2})(\tan \frac{A}{2}+\tan \frac{B}{2})\)

\(=\tan \frac{A}{2}.\tan \frac{B}{2}+\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{\tan (\frac{A+B}{2})}\)

\(=\tan \frac{A}{2}.\tan \frac{B}{2}+\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2}.\tan \frac{B}{2}}}\)

\(=\tan \frac{A}{2}.\tan \frac{B}{2}+1-\tan \frac{A}{2}.\tan \frac{B}{2}=1=\text{VP}\)

Ta có đpcm.

Cũng giống phần a, ta biết do ABC là tam giác nhọn nên

\(\tan A, \tan B, \tan C>0\)

Đặt \(\tan A=x, \tan B=y, \tan C=z\). Ta có: \(xy+yz+xz=1\)

Và \(T=x+y+z\)

\(\Rightarrow T^2=x^2+y^2+z^2+2(xy+yz+xz)\)

Theo hệ quả quen thuộc của BĐT Cauchy:

\(x^2+y^2+z^2\geq xy+yz+xz\)

\(\Rightarrow T^2\geq 3(xy+yz+xz)=3\)

\(\Rightarrow T\geq \sqrt{3}\Leftrightarrow T_{\min}=\sqrt{3}\)

Dấu bằng xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\Leftrightarrow \angle A=\angle B=\angle C=60^0\)

Câu a)

Ta sử dụng 2 công thức:

∙tan(180−α)=−tanα∙tan⁡(180−α)=−tan⁡α

∙tan(α+β)=tanα+tanβ1−tanα.tanβ∙tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡α.tan⁡β

Áp dụng vào bài toán:

VT=tanA+tanB+tanC=tanA+tanB+tan(180−A−B)VT=tan⁡A+tan⁡B+tan⁡C=tan⁡A+tan⁡B+tan⁡(180−A−B)

=tanA+tanB−tan(A+B)=tanA+tanB−tanA+tanB1−tanA.tanB=tan⁡A+tan⁡B−tan⁡(A+B)=tan⁡A+tan⁡B−tan⁡A+tan⁡B1−tan⁡A.tan⁡B

=(tanA+tanB)(1+11−tanA.tanB)=(tanA+tanB).−tanA.tanB1−tanA.tanB=(tan⁡A+tan⁡B)(1+11−tan⁡A.tan⁡B)=(tan⁡A+tan⁡B).−tan⁡A.tan⁡B1−tan⁡A.tan⁡B

=−tanA.tanB.tanA+tanB1−tanA.tanB=−tanA.tanB.tan(A+B)=−tan⁡A.tan⁡B.tan⁡A+tan⁡B1−tan⁡A.tan⁡B=−tan⁡A.tan⁡B.tan⁡(A+B)

=tanA.tanB.tan(180−A−B)=tan⁡A.tan⁡B.tan⁡(180−A−B)

=tanA.tanB.tanC=VP=tan⁡A.tan⁡B.tan⁡C=VP

Do đó ta có đpcm

Tam giác ABCABC có ba góc nhọn nên tanA,tanB,tanC>0tan⁡A,tan⁡B,tan⁡C>0

Áp dụng BĐT Cauchy ta có:

P=tanA+tanB+tanC≥33√tanA.tanB.tanCP=tan⁡A+tan⁡B+tan⁡C≥3tan⁡A.tan⁡B.tan⁡C3

⇔P=tanA+tanB+tanC≥33√tanA+tanB+tanC⇔P=tan⁡A+tan⁡B+tan⁡C≥3tan⁡A+tan⁡B+tan⁡C3

⇒P≥33√P⇒P≥3P3

⇒P3≥27P⇔P(P2−27)≥0⇒P3≥27P⇔P(P2−27)≥0

⇒P2−27≥0⇒P≥3√3⇒P2−27≥0⇒P≥33

Vậy Pmin=3√3Pmin=33. Dấu bằng xảy ra khi ∠A=∠B=∠C=600

Ta có:

\(\dfrac{tanA}{tan^3B}=\dfrac{tanA}{tanB}.\dfrac{1}{tan^2B}=\dfrac{\dfrac{sinA}{cosA}}{\dfrac{sinB}{cosB}}.\dfrac{cos^2B}{sin^2B}\)

\(=\dfrac{sinA}{sinB}.\dfrac{cosB}{cosA}.\dfrac{cos^2B}{sin^2B}\)

\(=\dfrac{a}{b}.\dfrac{\dfrac{a^2+c^2-b^2}{2ac}}{\dfrac{b^2+c^2-a^2}{2bc}}.\dfrac{\left(\dfrac{a^2+c^2-b^2}{2ac}\right)^2}{1-\left(\dfrac{a^2+c^2-b^2}{2ac}\right)^2}\)

\(=\dfrac{a^2+c^2-b^2}{b^2+c^2-a^2}.\dfrac{\left(a^2+c^2-b^2\right)^2}{\left(2ac\right)^2-\left(a^2+c^2-b^2\right)^2}\)

\(=\dfrac{\left(a^2+c^2-b^2\right)^3}{b^2+c^2-a^2}.\dfrac{1}{\left[\left(a+c\right)^2-b^2\right]\left[b^2-\left(a-c\right)^2\right]}\)

\(=\dfrac{\left(a^2+c^2-b^2\right)^3}{b^2+c^2-a^2}.\dfrac{1}{\left(a+b+c\right)\left(a+c-b\right)\left(b+c-a\right)\left(a+b-c\right)}\)

Biến đổi tương tự, ta có BĐT tương đương với BĐT đã cho:

\(\dfrac{\left(a^2+c^2-b^2\right)^3}{b^2+c^2-a^2}+\dfrac{\left(a^2+b^2-c^2\right)^3}{a^2+c^2-b^2}+\dfrac{\left(b^2+c^2-a^2\right)^3}{a^2+b^2-c^2}\ge\left(a+b+c\right)\left(b+c-a\right)\left(a+c-b\right)\left(a+b-c\right)\)

Ta có BĐT phụ sau:

\(\dfrac{x^3}{y}+\dfrac{y^3}{z}+\dfrac{z^3}{x}\ge xy+yz+xz\left(\text{*}\right)\) với \(x,y,z>0\)

Chứng minh:

Áp dụng BĐT cộng mẫu:

\(\dfrac{x^3}{y}+\dfrac{y^3}{z}+\dfrac{z^3}{x}=\dfrac{x^4}{xy}+\dfrac{y^4}{yz}+\dfrac{z^4}{xz}\)

\(\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{xy+yz+xz}\ge\dfrac{\left(xy+yz+xz\right)^2}{xy+yz+xz}=xy+yz+xz\)(đpcm)

Đẳng thức xảy ra khi và chỉ khi \(x=y=z\)

Áp dụng BĐT \(\left(\text{*}\right)\), với đk \(\Delta ABC\) có ba góc nhọn, ta có:

\(\dfrac{\left(a^2+c^2-b^2\right)^3}{b^2+c^2-a^2}+\dfrac{\left(a^2+b^2-c^2\right)^3}{a^2+c^2-b^2}+\dfrac{\left(b^2+c^2-a^2\right)^3}{a^2+b^2-c^2}\ge\left(a^2+c^2-b^2\right)\left(a^2+b^2-c^2\right)+\left(a^2+b^2-c^2\right)\left(b^2+c^2-a^2\right)+\left(b^2+c^2-a^2\right)\left(a^2+c^2-b^2\right)\)

Ta chứng minh được:

\(\left(a^2+c^2-b^2\right)\left(a^2+b^2-c^2\right)+\left(a^2+b^2-c^2\right)\left(b^2+c^2-a^2\right)+\left(b^2+c^2-a^2\right)\left(a^2+c^2-b^2\right)=\left(a+b+c\right)\left(b+c-a\right)\left(a+c-b\right)\left(a+b-c\right)\)

\(=-a^4-b^4-c^4+2a^2b^2+2b^2c^2+2a^2c^2\)

Vậy ta có BĐT cần chứng minh, đẳng thức xảy ra khi và chỉ khi \(\widehat{A}=\widehat{B}=\widehat{C}=60^0\)

lên mạng ghi cách đổi kg ra tấn là nó ra nha bn         

2 tấn 65kg = 2065kg

2065kg = 2 tấn = 0,65 tạ

đúng ko?

?????

\(2tan65kg=2tan0,65ta\)

chuc bn hoc gioi!

nhae@

hihi

856cm2;42dm2;2500000m2

18000m2;10m2;0,8m2

6m29dm2,2dm27cm2;300m2

rối mắt vãi

Nghia la no cu thang hang nhau do ban