\(\dfrac{120}{x}+\dfrac{120}{x-10}=\dfrac{3}{5}\left(dkxd:x>0,x\ne10\right)\)

\(\Leftrightarrow\dfrac{120}{x}+\dfrac{120}{x-10}-\dfrac{3}{5}=0\)

\(\Leftrightarrow\dfrac{120.5\left(x-10\right)+5.120x-3x\left(x-10\right)}{5x\left(x-10\right)}=0\)

\(\Leftrightarrow600x-6000+600x-3x^2+30x=0\)

\(\Leftrightarrow-3x^2+1230x-6000=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\approx405\\x\approx5\end{matrix}\right.\)\(\left(tmdk\right)\)

Vậy ...

\(1+\dfrac{1}{6}+\dfrac{120-x}{x}=\dfrac{120}{x}\)

\(1+\dfrac{1}{6}+\dfrac{126-\left(x+6\right)}{x+6}=\dfrac{120}{x}\)

\(1+\dfrac{1}{6}-1+\dfrac{126}{x+6}=\dfrac{120}{x}\)

\(\dfrac{1}{6}+\dfrac{126}{x+6}=\dfrac{120}{x}\)

\(\dfrac{126}{x+6}=\dfrac{120}{x}-\dfrac{1}{6}=\dfrac{120.6}{6x}-\dfrac{x}{6x}\)

\(\dfrac{126}{x+6}=\dfrac{126.6-x}{6x}\)

\(126.6.x=\left(126.6.-x\right)\left(x+6\right)\)ok

đk: x khác -6 ,làm toán là khôn khéo, bn tim msc vế trái =6(x+6)

có: (6(x+6) + (x+6) + 6(120-x)) /6(x+6) = 120/x

bây gio bn rut gon r cho tich trung tỷ = ngoai ty la tim dc x

\(\left\{{}\begin{matrix}x-y=10\\\dfrac{-120\left(x-y\right)}{xy}=\dfrac{2}{5}\end{matrix}\right.\) \(\Rightarrow\dfrac{-1200}{xy}=\dfrac{2}{5}\Rightarrow xy=-3000\)

Ta được hệ: \(\left\{{}\begin{matrix}x-y=10\\xy=-3000\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=y+10\\xy=-3000\end{matrix}\right.\)

Thay pt trên vào dưới:

\(\left(y+10\right).y=-3000\Rightarrow y^2+10y+3000=0\)

\(\Rightarrow\) pt vô nghiệm

Vậy hệ đã cho vô nghiệm

1) \(\dfrac{120\left(x-10\right)}{x\left(x-10\right)}-\dfrac{120x}{x\left(x-10\right)}=1\)

=> \(\dfrac{120x-1200-120x}{x\left(x-10\right)}=1\)

=> x(x-10)=-1200

=> x²-10x+1200=0

=> (x²-10x+25)+1175=0

=> (x-5)²+1175>0

=> pt vo nghiem

-90

\(\dfrac{10-x}{100}\) + \(\dfrac{20-x}{110}\)+\(\dfrac{30-x}{120}\)=3

<=> \(\dfrac{10-x}{100}\)-1+\(\dfrac{20-x}{110}\)-1+\(\dfrac{30-x}{120}\)-1 = 0

<=> \(\dfrac{-x-90}{100}\)+\(\dfrac{-x-90}{110}\)+\(\dfrac{-x-90}{120}\)=0

<=> (-x-90) ( \(\dfrac{1}{100}\)+\(\dfrac{1}{110}\)+\(\dfrac{1}{120}\))=0

<=> (-x-90) = 0 ( do 1/100 +1/110+1/120 khác 0)

<=> -x-90 = 0

<=> -x = 90

<=> x =-90

Vậy nghiệm của pt là x=-90

1.

\(\Leftrightarrow2sinx.cosx+2cosx=0\)

\(\Leftrightarrow2cosx\left(sinx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sinx=-1\end{matrix}\right.\)

\(\Leftrightarrow cosx=0\) (do \(cosx=0\Leftrightarrow sinx=\pm1\) bao hàm luôn cả pt \(sinx=-1\))

\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)

2.

\(\Leftrightarrow\left[{}\begin{matrix}2x-10^0=60^0+k360^0\\2x-10^0=120^0+n360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=35^0+k180^0\\x=65^0+n180^0\end{matrix}\right.\)

Do \(-120^0< x< 90^0\Rightarrow\left\{{}\begin{matrix}-120^0< 35^0+k180^0< 90^0\\-120^0< 65^0+n180^0< 90^0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}k=0\\n=\left\{-1;0\right\}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=35^0\\x=-115^0\\x=65^0\end{matrix}\right.\)

3. Làm tương tự câu 2

4.

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2}cos\left(10x+\dfrac{4\pi}{5}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}cos\left(\dfrac{x}{2}-2\pi\right)\right)=0\)

\(\Leftrightarrow cos\left(10x+\dfrac{4\pi}{5}\right)+cos\left(\dfrac{x}{2}-2\pi\right)=0\)

\(\Leftrightarrow cos\left(10x+\dfrac{4\pi}{5}\right)+cos\left(\dfrac{x}{2}\right)=0\)

\(\Leftrightarrow cos\left(10x+\dfrac{4\pi}{5}\right)=-cos\left(\dfrac{x}{2}\right)=cos\left(\pi-\dfrac{x}{2}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}10x+\dfrac{4\pi}{5}=\pi-\dfrac{x}{2}+k2\pi\\10x+\dfrac{4\pi}{5}=\dfrac{x}{2}-\pi+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)

a: \(\Leftrightarrow x^2+x-6+2x-6=10x-20+50\)

\(\Leftrightarrow x^2+3x-12-10x-30=0\)

\(\Leftrightarrow x^2-7x-42=0\)

\(\text{Δ}=\left(-7\right)^2-4\cdot1\cdot\left(-42\right)=217>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{7-\sqrt{217}}{2}\\x_2=\dfrac{7+\sqrt{217}}{2}\end{matrix}\right.\)

b: \(\Leftrightarrow x^2-3x+5=-x^2+4\)

\(\Leftrightarrow2x^2-3x+1=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)

hay \(x\in\left\{\dfrac{1}{2};1\right\}\)

x^2-3x+2=(x-1)(x-2)

dk x≠1;2

1+(x-5)(x-1)=3/10(x^2-3x+2)

10+10x^2-60x+50=3x^2-9x+6

7x^2-54x-54=0

x=(27±3√123)/7

\(\dfrac{1}{x^2-3x+2}-\dfrac{x-5}{2-x}=\dfrac{3}{10}\)

⇔ \(\dfrac{1}{x^2-x-2x+2}+\dfrac{x-5}{x-2}=\dfrac{3}{10}\)

⇔ \(\dfrac{10}{10\left(x-1\right)\left(x-2\right)}+\dfrac{10\left(x-5\right)\left(x-1\right)}{10\left(x-1\right)\left(x-2\right)}=\dfrac{3\left(x^2-3x+2\right)}{10\left(x-1\right)\left(x-2\right)}\)( x # 1 ; x # 2)

⇔ 10 + 10( x² - 6x + 5)= 3(x² - 3x + 2)

⇔ 10 + 10x² - 60x + 50 = 3x² - 9x + 6

⇔ 7x² - 51x - 54 = 0

Phân tích ra

\(\dfrac{1}{120}\cdot120+x:\dfrac{1}{3}=-4\)

\(\Leftrightarrow1+x\cdot3=-4\)

\(\Leftrightarrow3x=-5\)

\(\Leftrightarrow x=-\dfrac{5}{3}\)

\(\dfrac{1}{120}.120+x:\dfrac{1}{3}=-4\) 

            \(1+x:\dfrac{1}{3}=-4\) 

                  \(x:\dfrac{1}{3}=-4-1\) 

                  \(x:\dfrac{1}{3}=-5\) 

                        \(x=-5.\dfrac{1}{3}\) 

                        \(x=\dfrac{-5}{3}\)