8: =>6x^2-9x+2x-3-6x^2-12x=16

=>-19x=19

=>x=-1

Đáp số của bài toán đúng nhưng lời giải của bạn Hà chưa đầy đủ.

Lời giải của bạn Hà thiếu bước tìm điều kiện xác định và bước đối chiếu giá trị của x tìm được với điều kiện để kết luận nghiệm.

Trong bài toán trên thì điều kiện xác định của phương trình là:

x ≠ - 3/2 và x  ≠  - 1/2

So sánh với điều kiện xác định thì giá trị x = - 4/7 thỏa mãn.

Vậy x = - 4/7 là nghiệm của phương trình.

\(\frac{1}{2x^2+10x+12}+\frac{1}{2x^2+14x+24}+\frac{1}{2x^2+18x+40}+\frac{1}{2x^2+22x+60}=\frac{1}{8}\)

<=> \(\frac{1}{2x^2+6x+4x+12}+\frac{1}{2x^2+6x+8x+24}+\frac{1}{2x^2+8x+10x+40}+\frac{1}{2x^2+12x+10x+60}=\frac{1}{8}\)

<=> \(\frac{1}{2x\left(x+3\right)+4\left(x+3\right)}+\frac{1}{2x\left(x+3\right)+8\left(x+3\right)}+\frac{1}{2x\left(x+4\right)+10\left(x+4\right)}+\frac{1}{2x\left(x+6\right)+10\left(x+6\right)}=\frac{1}{8}\)

<=> \(\frac{1}{\left(x+3\right)\left(2x+4\right)}+\frac{1}{\left(x+3\right)\left(2x+8\right)}+\frac{1}{\left(x+4\right)\left(2x+10\right)}+\frac{1}{\left(x+6\right)\left(2x+10\right)}=\frac{1}{8}\)

<=> \(\frac{1}{2\left(x+2\right)\left(x+3\right)}+\frac{1}{2\left(x+3\right)\left(x+4\right)}+\frac{1}{2\left(x+4\right)\left(x+5\right)}+\frac{1}{2\left(x+5\right)\left(x+6\right)}=\frac{1}{8}\)

<=> \(\frac{1}{2}.\left[\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}\right]=\frac{1}{8}\)

<=> \(\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}=\frac{1}{8}:\frac{1}{2}\)

<=> \(\frac{1}{x+2}-\frac{1}{x+6}=\frac{1}{4}\)

<=> \(\frac{4\left(x+6\right)-4\left(x+2\right)}{4\left(x+2\right)\left(x+6\right)}=\frac{\left(x+2\right)\left(x+6\right)}{4\left(x+2\right)\left(x+6\right)}\)

<=> \(4\left(x+6\right)-4\left(x+2\right)=\left(x+2\right)\left(x+6\right)\)

<=> \(4\left(x+6-x-2\right)=x^2+8x+12\)

<=> \(4.4=x^2+8x+12\)

<=> \(x^2+8x-4=0\)

<=> ...

Đến đây bạn tự giải tiếp. Mình bấm máy 570ES PLUS II thì ra nghiệm \(x\approx0,47\).

a) \(\left|-\frac{2}{11}+\frac{3}{22}x\right|-\frac{1}{2}=\frac{5}{7}\)

=> \(\left|-\frac{2}{11}+\frac{3}{22}x\right|=\frac{17}{14}\)

=> \(\orbr{\begin{cases}-\frac{2}{11}+\frac{3}{22}x=\frac{17}{14}\\-\frac{2}{11}+\frac{3}{22}x=-\frac{17}{14}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{215}{21}\\x=-\frac{53}{7}\end{cases}}\)

b) \(-\frac{7}{8}x-5\frac{3}{4}=3\)

=> \(-\frac{7}{8}x-\frac{23}{4}=3\)

=> \(-\frac{7}{8}x=3+\frac{23}{4}=\frac{35}{4}\)

=> \(x=\frac{35}{4}:\left(-\frac{7}{8}\right)=\frac{35}{4}\cdot\left(-\frac{8}{7}\right)=-10\)

c) \(2x+\left(-\frac{2}{7}\right)-7=-11\)

=> \(2x-\frac{2}{7}-7=-11\)

=> \(2x=-11+7+\frac{2}{7}=-\frac{26}{7}\)

=> \(x=\left(-\frac{26}{7}\right):2=-\frac{13}{7}\)

d) \(\frac{3}{7}+x:\frac{14}{15}=\frac{1}{2}\)

=> \(x:\frac{14}{15}=\frac{1}{2}-\frac{3}{7}=\frac{1}{14}\)

=> \(x=\frac{1}{14}\cdot\frac{14}{15}=\frac{1}{15}\)

c: \(\cot\left(\frac{\pi}{4}-\frac{x}{2}\right)\)

\(=\tan\left(\frac{\pi}{2}-\frac{\pi}{4}+\frac{x}{2}\right)=\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)=\frac{\tan\left(\frac{\pi}{4}\right)+\tan\left(\frac{x}{2}\right)}{1-\tan\left(\frac{\pi}{4}\right)\cdot\tan\left(\frac{x}{2}\right)}=\frac{\tan\left(\frac{x}{2}\right)+1}{1-\tan\left(\frac{x}{2}\right)}\)

\(=\left(\frac{\sin\left(\frac{x}{2}\right)}{cos\left(\frac{x}{2}\right)}+1\right):\left(1-\frac{\sin\left(\frac{x}{2}\right)}{cos\left(\frac{x}{2}\right)}\right)=\frac{\sin\left(\frac{x}{2}\right)+cos\left(\frac{x}{2}\right)}{cos\left(\frac{x}{2}\right)-\sin\left(\frac{x}{2}\right)}\) (1)

\(\frac{cosx}{1-\sin x}=\frac{cos\left(2\cdot\frac{x}{2}\right)}{1-\sin\left(2\cdot\frac{x}{2}\right)}=\frac{cos^2\left(\frac{x}{2}\right)-\sin^2\left(\frac{x}{2}\right)}{\left(cos\left(\frac{x}{2}\right)-\sin\left(\frac{x}{2}\right)\right)^2}\) (2)

Từ (1),(2) suy ra \(\frac{cosx}{1-\sin x}=\cot\left(\frac{\pi}{4}-\frac{x}{2}\right)\)

b: \(\tan\left(x+\frac{\pi}{4}\right)=\frac{\tan x+\tan\left(\frac{\pi}{4}\right)}{1-tanx\cdot\tan\left(\frac{\pi}{4}\right)}=\frac{\tan x+1}{1-\tan x}\)

\(=\left(\frac{\sin x}{cosx}+1\right):\left(1-\frac{\sin x}{cosx}\right)=\frac{\sin x+cosx}{cosx-\sin x}\)

\(\frac{1+\sin2x}{cos2x}=\frac{1+2\cdot\sin x\cdot cosx}{cos^2x-\sin^2x}=\frac{\left(cosx+\sin x\right)^2}{\left(cosx-\sin x\right)\left(cosx+\sin x\right)}=\frac{cosx+\sin x}{cosx-\sin x}\)

Do đó: \(\tan\left(x+\frac{\pi}{4}\right)=\frac{1+\sin2x}{cos2x}\)

1.Pt \(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=sin\left(x+\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=cos\left(\dfrac{\pi}{6}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}-x+k2\pi\\2x-\dfrac{\pi}{3}=x-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)

\(\Rightarrow x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\)\(\left(k\in Z\right)\)

2.\(sin^22x+cos^23x=1\)

\(\Leftrightarrow\dfrac{1-cos4x}{2}+\dfrac{1+cos6x}{2}=1\)

\(\Leftrightarrow cos6x=cos4x\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{k\pi}{5}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow x=\dfrac{k\pi}{5}\)\(\left(k\in Z\right)\) (Gộp nghiệm)

Vậy...

3. \(Pt\Leftrightarrow\left(sinx+sin3x\right)+\left(sin2x+sin4x\right)=0\)

\(\Leftrightarrow2.sin2x.cosx+2.sin3x.cosx=0\)

\(\Leftrightarrow2cosx\left(sin2x+sin3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin3x=-sin2x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\sin3x=sin\left(\pi+2x\right)\end{matrix}\right.\)(\(k\in Z\))

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\pi+k2\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\)(\(k\in Z\))\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\) (\(k\in Z\))

Vậy...

4. Pt\(\Leftrightarrow\dfrac{1-cos2x}{2}+\dfrac{1-cos4x}{2}=\dfrac{1-cos6x}{2}\)

\(\Leftrightarrow cos2x+cos4x=1+cos6x\)

\(\Leftrightarrow2cos3x.cosx=2cos^23x\)

\(\Leftrightarrow\left[{}\begin{matrix}cos3x=0\\cosx=cos3x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=-k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)

Vậy...