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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 10: tan 2x = 24/7 tính Q= sinx-cosx

Toán Lớp 10: tan 2x = 24/7
tính Q= sinx-cosx

Comments ( 2 )

  1. Giải đáp:
    $Q\in\left\{-\dfrac{7}{5};-\dfrac{1}{5};\dfrac{1}{5};\dfrac{7}{5}\right\}$
    Lời giải và giải thích chi tiết:
    $\tan2x=\dfrac{24}{7}$
    $⇒\dfrac{2\tan x}{1-\tan^2x}=\dfrac{24}{7}$
    $⇒14\tan x=24-24\tan^2x$
    $⇒12\tan^2x+7\tan x-12=0$
    $⇒\left[ \begin{array}{l}\tan x=\dfrac{3}{4}\\\tan x=-\dfrac{4}{3}\end{array} \right.⇒\left[ \begin{array}{l}\dfrac{\sin x}{\cos x}=\dfrac{3}{4}\\\dfrac{\sin x}{\cos x}=-\dfrac{4}{3}\end{array} \right.⇒\left[ \begin{array}{l}\cos x=\dfrac{4}{3}\sin x\\\cos x=-\dfrac{3}{4}\sin x\end{array} \right.$
    TH1: $\cos x=\dfrac{4}{3}\sin x$
    $⇒\sin^2x+\cos^2x=1$
    $⇒\sin^2x+\dfrac{16}{9}\sin^2x=1$
    $⇒\dfrac{25}{9}\sin^2x=1$
    $⇒\sin^2x=\dfrac{9}{25}$
    $⇒\sin x=±\dfrac{3}{5}$
    $⇒\cos x=±\dfrac{4}{5}$
    $⇒Q=\sin x-\cos x=\mp\dfrac{1}{5}$
    TH2: $\cos x=-\dfrac{3}{4}\sin x$
    $⇒\sin^2x+\cos^2x=1$
    $⇒\sin^2x+\dfrac{9}{16}\sin^2x=1$
    $⇒\dfrac{25}{16}\sin^2x=1$
    $⇒\sin^2x=\dfrac{16}{25}$
    $⇒\sin x=±\dfrac{4}{5}$
    $⇒\cos x=\mp\dfrac{3}{5}$
    $⇒Q=\sin x-\cos x=±\dfrac{7}{5}$
    Vậy $Q\in\left\{-\dfrac{7}{5};-\dfrac{1}{5};\dfrac{1}{5};\dfrac{7}{5}\right\}$.

  2. Giải đáp:Tan2x= 24/7 ⇔Cot2x=7/24
    (1) ADCT:
    1/Sin²2x =1+(7/24)²= 625/576
    ⇒Sin²2x= 576/625
    ⇒Sin2x=±24/25
    Q= Sinx – Cosx
    ⇒Q² =Sin²x + Cos²x -2SinxCosx
    = 1-Sin2x
    = 1-24/25=1/25
    ⇒Q =±1/5
    với Sin2x = -24/25
    Q²=1+24/25 = 49/25
    ⇒ Q= ±7/5
    xin 5s và ctlhn nha
     

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222-9+11+12:2*14+14 = ? ( )