\(\tan A+tanB=\dfrac{sinC}{cosAcosB}\left(vớiA,B,C\right)làbacạnhcủatamgiacABC\)giúp mình với hic
Cho A, B, C là 3 góc nhọn của tam giác ABC. Chứng minh:
a) \(tanA+tanB+tanC=tanA.tanB.tanC\)
Tính min P với \(P=tanA+tanB+tanC\)
b) \(tan\left(\dfrac{A}{2}\right).tan\left(\dfrac{B}{2}\right)+tan\left(\dfrac{B}{2}\right)tan\left(\dfrac{C}{2}\right)+tan\left(\dfrac{C}{2}\right).tan\left(\dfrac{A}{2}\right)=1\)
Tìm min T với \(T=tan\left(\dfrac{A}{2}\right)+tan\left(\dfrac{B}{2}\right)+tan\left(\dfrac{C}{2}\right)\)
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)=−tanα
∙tan(α+β)=tanα+tanβ1−tanα.tanβ∙tan(α+β)=tanα+tanβ1−tanα.tanβ
Áp dụng vào bài toán:
VT=tanA+tanB+tanC=tanA+tanB+tan(180−A−B)VT=tanA+tanB+tanC=tanA+tanB+tan(180−A−B)
=tanA+tanB−tan(A+B)=tanA+tanB−tanA+tanB1−tanA.tanB=tanA+tanB−tan(A+B)=tanA+tanB−tanA+tanB1−tanA.tanB
=(tanA+tanB)(1+11−tanA.tanB)=(tanA+tanB).−tanA.tanB1−tanA.tanB=(tanA+tanB)(1+11−tanA.tanB)=(tanA+tanB).−tanA.tanB1−tanA.tanB
=−tanA.tanB.tanA+tanB1−tanA.tanB=−tanA.tanB.tan(A+B)=−tanA.tanB.tanA+tanB1−tanA.tanB=−tanA.tanB.tan(A+B)
=tanA.tanB.tan(180−A−B)=tanA.tanB.tan(180−A−B)
=tanA.tanB.tanC=VP=tanA.tanB.tanC=VP
Do đó ta có đpcm
Tam giác ABCABC có ba góc nhọn nên tanA,tanB,tanC>0tanA,tanB,tanC>0
Áp dụng BĐT Cauchy ta có:
P=tanA+tanB+tanC≥33√tanA.tanB.tanCP=tanA+tanB+tanC≥3tanA.tanB.tanC3
⇔P=tanA+tanB+tanC≥33√tanA+tanB+tanC⇔P=tanA+tanB+tanC≥3tanA+tanB+tanC3
⇒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
Chứng minh rằng \(\dfrac{a}{bc}+\dfrac{b}{ac}+\dfrac{c}{ab}\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)vớia,b,c>0\)
Giups mình với !!!!!!!!!!!!!!!!!!!!!!
\(\dfrac{a}{bc}+\dfrac{b}{ac}>=2\cdot\sqrt{\dfrac{a}{bc}\cdot\dfrac{b}{ac}}=\dfrac{2}{cc}\)
\(\dfrac{b}{ca}+\dfrac{c}{ab}>=2\cdot\sqrt{\dfrac{bc}{ca\cdot ab}}=\dfrac{2}{a}\)
\(\dfrac{c}{ab}+\dfrac{a}{bc}>=2\cdot\sqrt{\dfrac{a\cdot c}{a\cdot b\cdot c\cdot b}}=\dfrac{2}{b}\)
=>a/bc+b/ac+c/ab>=2(1/a+1/b+1/c)
Bạn nào giúp mình vs nhá:===thanks mọi người nhiều lắm^^
1/ cho tam giác ABC. cmr:
\(\dfrac{1}{sinA}+\dfrac{1}{sinB}+\dfrac{1}{sinC}=\dfrac{1}{2}.\left(tan\dfrac{A}{2}+tan\dfrac{B}{2}+tan\dfrac{C}{2}+cot\dfrac{A}{2}.cot\dfrac{B}{2}.cot\dfrac{C}{2}\right)\)
2,cmr:
\(\left(a-b\right)tan\dfrac{A}{2}.tan\dfrac{B}{2}+\left(b-c\right)tan\dfrac{B}{2}.tan\dfrac{C}{2}+\left(c-a\right)tan\dfrac{C}{2}.tan\dfrac{A}{2}=0\)
giúp mình với
A=\(\dfrac{1-0,5\cdot\left(3,84-2,4\right):0,8}{\dfrac{4}{5}-\left(1\dfrac{1}{3}-2\dfrac{1}{6}\right)-1,5}\)
mình cảm ơn
Ta có: \(A=\dfrac{1-0.5\cdot\left(3.84-2.4\right):0.8}{\dfrac{4}{5}-\left(1\dfrac{1}{3}-2\dfrac{1}{6}\right)-1.5}\)
\(=\dfrac{1-0.5\cdot1.44:0.8}{\dfrac{4}{5}-\left(\dfrac{4}{3}-\dfrac{13}{6}\right)-\dfrac{3}{2}}\)
\(=\dfrac{1-0.9}{\dfrac{4}{5}+\dfrac{5}{6}-\dfrac{3}{2}}=\dfrac{0.1}{\dfrac{2}{15}}=\dfrac{3}{4}\)
Giải:
A=1-0,5.(3,84-2,4):0,8 / 4/5-(1 1/3 - 2 1/6)-1,5
A=1-0,5.1,44:0.8 / 4/5-(4/3-13/6)-3/2
A=1-0.9 / 4/5-(-5/6)-3/2
A=0.1 / 2/15
A= 1/10 : 2/15
A=3/4
Chúc bạn học tốt!
a)\(\sqrt{4\left(a-3\right)^2}vớia\ge3\)
b)\(\sqrt{a^2\left(a+1\right)^2}vớia>0\)
c)\(\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}vớia< 0,b\ne0\)
a) \(\sqrt{4\left(a-3\right)^2}=2\left(a-3\right)=2a-6\)
b) \(\sqrt{a^2\left(a+1\right)^2}=a\left(a+1\right)=a^2+a\)
c) \(\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\dfrac{1}{\sqrt{8}\left|a\right|}=\dfrac{1}{-\sqrt{8}a}=\dfrac{-\sqrt{8}}{8a}\)
a: \(\sqrt{4\left(a-3\right)^2}=2\cdot\left(a-3\right)=2a-6\)
b: \(\sqrt{a^2\left(a+1\right)^2}=a\left(a+1\right)=a^2+a\)
c: \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\sqrt{\dfrac{2}{16a^2}}=-\dfrac{\sqrt{2}}{4a}\)
Chứng minh rằng trong tam giác ABC có:
a) tanB = tan( A+C)
b) sinC = sin( A +B)
c) cos A = -cos (B+C)
a: ΔABC có góc B+góc C+góc A=180 độ
=>góc B=180 độ-góc C-góc A
=>tan B=tan(A+C)
b: ΔABC có góc C+góc B+góc A=180 độ
=>góc C=180 độ-góc B-góc A
=>sin C=sin(A+B)
c: Xét ΔABC có góc A+góc B+góc C=180 độ
=>góc A=180 độ-góc B-góc C
=>cosA=-cos(B+C)
Cho tam giác abc. Chứng minh rằng: tan\(\left(\dfrac{B+C}{2}\right)\)= cot\(\left(\dfrac{A}{2}\right)\)
mọi người giúp mình với ạ nếu đc có thể giải thích giúp mình luôn đc ko
Ta có:
\(\widehat{A}+\widehat{B}+\widehat{C}=180^o\) (tính chất tổng 3 góc trong 1 tam giác)
\(\Rightarrow\dfrac{\widehat{A}+\widehat{B}+\widehat{C}}{2}=90^o\)
\(\Rightarrow\dfrac{\widehat{B}+\widehat{C}}{2}=90^o-\dfrac{\widehat{A}}{2}\)
\(\Rightarrow\)\(tan\left(\dfrac{\widehat{B}+\widehat{C}}{2}\right)=tan\left(90^o-\widehat{\dfrac{A}{2}}\right)\)
\(\Rightarrow tan\left(\dfrac{\widehat{B}+\widehat{C}}{2}\right)=cot\dfrac{A}{2}\)
rút gọn các biểu thức sau
A=\(\frac{tan\alpha+tanb}{tan\left(a+b\right)}-\frac{tan\alpha-tanb}{tan\left(a-b\right)}\)
B=\(\frac{cos^3x-cos3x}{cosx}+\frac{sin^3+sin3x}{sinx}\)
\(A=\frac{tana+tanb}{tan\left(a+b\right)}-\frac{tana-tanb}{tan\left(a-b\right)}=\frac{tana+tanb}{\frac{tana+tanb}{1-tana\cdot tanb}}-\frac{tana-tanb}{\frac{tana-tanb}{1+tana\cdot tanb}}\\ \Leftrightarrow A=1-tana\cdot tanb-1-tana\cdot tanb=-2tana\cdot tanb\)
\(B=\frac{cos^3x-cos3x}{cosx}+\frac{sin^3x+sin3x}{sinx}\\ B=\frac{cos^3x-4cos^3x+3cosx}{cosx}+\frac{sin^3x+3sinx-4sin^3x}{sinx}\\ B=\frac{-3cos^3x+3cosx}{cosx}+\frac{-3sin^3x+3sinx}{sinx}\\ B=\frac{cosx\left(-3cos^2x+3\right)}{cosx}+\frac{sinx\left(-3sin^2x+3\right)}{sinx}\\ B=-3cos^2x+3-3sin^2x+3=6-3\left(sin^2x+cos^2x\right)=6-3=3\)
~~~~~~~~chúc bạn học tốt~~~~~~~~~~
tính GTBT:
a)a.\(\left(\dfrac{-3}{2}\right)+a.\dfrac{1}{4}-a.\dfrac{5}{6}vớia=\dfrac{3}{5}\)
b)\(\dfrac{2}{5}.b-\dfrac{1}{3}.b+b.\left(\dfrac{-1}{2}\right)vớib=\dfrac{6}{13}\)
c)c.\(\dfrac{5}{6}+\dfrac{3}{4}.c-\dfrac{11}{12}.c\) với c=\(\dfrac{2019}{2020}\)
a) Ta có: \(a\left(-\dfrac{3}{2}\right)+a\cdot\dfrac{1}{4}-a\cdot\dfrac{5}{6}\)
\(=a\left(-\dfrac{3}{2}+\dfrac{1}{4}-\dfrac{5}{6}\right)\)
\(=a\left(\dfrac{-18}{12}+\dfrac{3}{12}-\dfrac{10}{12}\right)\)
\(=a\cdot\dfrac{-25}{12}\)(1)
Thay \(a=\dfrac{3}{5}\) vào biểu thức (1), ta được:
\(\dfrac{3}{5}\cdot\dfrac{-25}{12}=\dfrac{-75}{60}=\dfrac{-5}{4}\)