ta có : 

\(\frac{x-1009}{1001}-1+\frac{x-4}{1003}-2+\frac{x+2010}{1005}-4=0\)

hay \(\frac{x-2010}{1001}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\Leftrightarrow x-2010=0\)

hay x =2010

Vậy phương trình có nghiệm x = 2010

a, Xét tam giác AHB và tam giác CAB có 

^B _ chung 

^AHB = ^CAB = 90⁰

Vậy tam giác AHB ~ tam giác CAB (g.g) 

\(\dfrac{HB}{AB}=\dfrac{AB}{BC}\Rightarrow AB^2=HB.BC=HB\left(HB+HC\right)=225\Rightarrow AB=15cm\)

-> HB + HC = 25 cm 

Theo định lí Pytago tam giác AHB vuông tại H

\(AH=\sqrt{AB^2-HB^2}=12cm\)

Theo định lí Pytago tam giác ABC vuông tại A

\(AC=\sqrt{BC^2-AB^2}=20cm\)

b, Xét tam giác AHB và tam giác CHA có 

^AHB = ^CHA = 90⁰

^HAB = ^HCA ( cùng phụ ^HAC ) 

Vậy tam giác AHB ~ tam giác CHA (g.g) 

\(\dfrac{AH}{CH}=\dfrac{HB}{HA}\Rightarrow AH^2=HB.HC\)

AB^2 = BH . BC (cma) 

`Answer:`

a. \(x^3+x^2-x+2=4x-1\)

\(\Leftrightarrow x^3+x^2-x-4x+2+1=0\)

\(\Leftrightarrow x^3+x^2-5x+3=0\)

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

\(\Leftrightarrow x^2.\left(x-1\right)+2x\left(x-1\right)-3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^2+3x-x-3\right)\left(x-1\right)=0\)

\(\Leftrightarrow[x\left(x+3\right)-\left(x+3\right)]\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2.\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)

b. \(x^4+35x^2-74=0\)

\(\Leftrightarrow x^4+37x^2-2x^2-74=0\)

\(\Leftrightarrow x^2.\left(x^2+37\right)-2\left(x^2+37\right)=0\)

\(\Leftrightarrow\left(x^2-2\right)\left(x^2+37\right)=0\)

Mà \(x^2+37\ne0\forall x\)

\(\Rightarrow x^2-2=0\)

\(\Leftrightarrow x^2=2\)

\(\Leftrightarrow x=\pm\sqrt{2}\)

c. \(2x^4+2x^3-76x^2+4x+8=0\)

\(\Leftrightarrow2\left(x^4+x^3-38x^2+2x+4\right)=0\)

\(\Leftrightarrow x^4+x^3-38x^2+2x+4=0\)

\(\Leftrightarrow x^4+7x^3+2x^2-6x^3-42x^2-12x+2x^2+14x+4=0\)

\(\Leftrightarrow x^2.\left(x^2+7x+2\right)-6x\left(x^2+7x+2\right)+2.\left(x^2+7x+2\right)=0\)

\(\Leftrightarrow\left(x^2-6x+2\right)\left(x^2+7x+2\right)=0\)

Trường hợp 1: \(x^2-6x+2\)

Ta có \(\Delta'=\left(-3\right)^2-2=7>0\)

Vậy phương trình có hai nghiệm phân biệt:

\(x_1=3-\sqrt{7}\)

\(x_2=3+\sqrt{7}\)

Trường hợp 2: \(x^2+7x+2=0\)

Ta có \(\Delta=7^2-4.2=41>0\)

Vậy phương trình có hai nghiệm phân biệt:

\(x_1=\frac{-7-\sqrt{41}}{2}\)

\(x_2=\frac{-7+\sqrt{41}}{2}\)

d. \(\left(3x+3\right)^4+\left(3x+5\right)^4=0\)

Mà \(\left(3x+3\right)^4\ge0;\left(3x+5\right)^4\ge0\forall x\inℝ\)

`=>` Để phương trình có nghiệm thì \(\hept{\begin{cases}3x+3=0\\3x+5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}3x=-3\\3x=-5\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\x=-\frac{5}{3}\end{cases}}\text{(Vô lý)}\)

Vậy phương trình vô nghiệm.

e. \(\left(2x+1\right)\left(2x+3\right)\left(2x+5\right)\left(2x+7\right)=169\)

\(\Leftrightarrow[\left(2x+1\right)\left(2x+7\right)][\left(2x+3\right)\left(2x+5\right)]-169=0\)

\(\Leftrightarrow\left(4x^2+16x+6+1\right)\left(4x^2+16x+6-1\right)-169=0\)

Đặt \(a=4x^2+16x+6\)

\(\Rightarrow\left(a+1\right)\left(a-1\right)-169=0\)

\(\Leftrightarrow a^2-1^2-169=0\)

\(\Leftrightarrow a^2-170=0\)

\(\Leftrightarrow a=\pm\sqrt{170}\)

\(\Rightarrow4x^2+16+6=\pm\sqrt{170}\)

\(\Leftrightarrow4x^2+16+16-10=\pm\sqrt{170}\)

\(\Leftrightarrow4\left(x+2\right)^2=\pm\sqrt{170}+10\)

\(\Leftrightarrow\left(x+2\right)^2=\left(\pm\sqrt{\frac{\pm\sqrt{170}+10}{4}}\right)^2\)

\(\Leftrightarrow x+2=\pm\sqrt{\frac{\pm\sqrt{170}+10}{4}}\)

\(\Leftrightarrow x=\pm\sqrt{\frac{\pm\sqrt{170}+10}{4}}-2\)

\(Tacó:\) \(x+y=3\)

\(\dfrac{x}{2}+\dfrac{x}{2}+y=3\)

Áp dụng BĐT Cô si cho 3 số, ta có:

\(3=\left(\dfrac{x}{2}+\dfrac{x}{2}+y\right)\ge3.^3\sqrt{\dfrac{x}{2}.\dfrac{x}{2}.y}=3.^3\sqrt{\dfrac{x^2y}{4}}\)

⇒\(1\ge^3\sqrt{\dfrac{x^2y}{4}}\)

⇒\(1\ge\dfrac{x^2y}{4}\Leftrightarrow x^2y\le4\)

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\(\left(1+\frac{1}{x}\right).\left(1+\frac{1}{y}\right).\left(1+\frac{1}{z}\right)=2\)

Giả sử \(x\ge y\ge z>0\)

\(\Rightarrow\frac{1}{x}\le\frac{1}{y}\le\frac{1}{z}\)

\(\Rightarrow1+\frac{1}{x}\le1+\frac{1}{y}\le1+\frac{1}{z}\)

\(\Rightarrow\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right)\le \left(1+\frac{1}{z}\right)^3\)

\(\Rightarrow2\le\left(1+\frac{1}{z}\right)^3\)

\(\Rightarrow1+\frac{1}{z}\ge\sqrt[3]{2}\)
\(\Rightarrow\frac{1}{z}\ge\sqrt[3]{2}-1\)

\(\Rightarrow z\le\frac{1}{\sqrt[3]{2}-1}< 4\)

Mà z thuộc N^* \(\Rightarrow z\in\left\{1;2;3\right\}\)

TH1 : \(z=1\)

\(\Rightarrow\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{1}\right)=2\)

\(\Rightarrow\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=1\)

Ta có : \(1+\frac{1}{x}>1;1+\frac{1}{y}>1\)\(\Rightarrow\left(\frac{1}{x}+1\right)\left(1+\frac{1}{y}\right)>1\left(lọai\right)\)

TH2 : \(z=2\)

\(\Rightarrow\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{2}\right)=2\)

\(\Rightarrow\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=\frac{4}{3}\)

Ta có : \(\left(1+\frac{1}{y}\right)^2\ge\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=\frac{4}{3}\)

\(\Rightarrow1+\frac{1}{y}\ge\sqrt{\frac{4}{3}}\)

\(\Rightarrow\frac{1}{y}\ge\frac{2\sqrt{3}}{3}-1\)

\(\Rightarrow y\le\frac{1}{\frac{2\sqrt{3}}{3}-1}< 7\)

\(\Rightarrow y\in\left\{1;2;3;4;5;6\right\}\)

Nếu y = 1 \(\Rightarrow\left(1+1\right)\left(1+\frac{1}{x}\right)=\frac{4}{3}\)

= > x = -3 ( loại )

Nếu y = 2 \(\Rightarrow\left(1+\frac{1}{2}\right)\left(1+\frac{1}{x}\right)=\frac{4}{3}\)

= > x = -9 ( loại )

Nếu y = 3 \(\Rightarrow\left(1+\frac{1}{3}\right)\left(1+\frac{1}{x}\right)=\frac{4}{3}\)

= > \(x\in\varnothing\)

Nếu y = 4 \(\Rightarrow\left(1+\frac{1}{4}\right)\left(1+\frac{1}{x}\right)=\frac{4}{3}\)

= > x = 15 ( tm )

Nếu y = 5 \(\Rightarrow\left(1+\frac{1}{5}\right)\left(1+\frac{1}{x}\right)=\frac{4}{3}\)

= > x = 9 ( tm )

Nếu y = 6 \(\Rightarrow\left(1+\frac{1}{6}\right)\left(1+\frac{1}{x}\right)=\frac{4}{3}\)

= > x = 7 ( tm )

TH3 : z =3 thì bạn làm tương tự nhé

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ĐKXĐ : \(x\ne-2,x\ne2\)

\(\frac{x+1}{x-2}+\frac{5}{2+x}=\frac{12}{x^2-4}+1\)

\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{x^2-4}-\frac{5\left(x-2\right)}{x^2-4}=\frac{12}{x^2-4}+\frac{x^2-4}{x^2-4}\)

\(\Leftrightarrow x^2+3x+2-5x+10=8+x^2\)

\(\Leftrightarrow12-2x=8\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\left(ktm\right)\)

Vậy phương trình vô nghiệm

\(\frac{x+1}{x-2}+\frac{5}{2+x}=\frac{12}{x^2-4}+1\)

ĐKXĐ: \(x\ne\pm2\)

\(\Leftrightarrow\frac{x+1}{x-2}+\frac{-5}{x-2}=\frac{12}{\left(x-2\right)\left(x+2\right)}+1\)

\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{-5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{12}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow\left(x+1\right)\left(x+2\right)-5\left(x+2\right)=12+\left(x-2\right)\left(x+2\right)\)

\(\Leftrightarrow x^2+2x+x+2-5x-10=12+x^2+2x-2x-4\)

\(\Leftrightarrow x^2-2x-8=8+x^2\)

\(\Leftrightarrow x^2-2x-8-8-x^2=0\)

\(\Leftrightarrow-2x-16=0\)

\(\Leftrightarrow-2x=0+16\)

\(\Leftrightarrow-2x=16\)

\(\Leftrightarrow x=16:\left(-2\right)\)

\(\Leftrightarrow x=-8\)(TMĐKXĐ)

Vậy S = \(\left\{-8\right\}\)

a, Ta có \(\orbr{\begin{cases}2x-3=1\\2x-3=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\left(loai\right)\\x=1\end{cases}}\)

Thay x = 1 ta được \(A=\frac{1+1}{2-1}=2\)

b, đk x khác -2 ; -1 

\(B=\frac{2x}{x+1}+\frac{3}{x-2}-\frac{2x^2+1}{x^2-x-2}\)

\(=\frac{2x^2-4x+3x+3-2x^2-1}{\left(x+1\right)\left(x-2\right)}=-\frac{1}{x+1}\)

c, \(P=AB=\frac{-x\left(x+1\right)}{\left(x+1\right)\left(2-x\right)}=\frac{x}{x-2}=\frac{x-2+2}{x-2}=1+\frac{2}{x-2}\)

`Answer:`

\(\frac{x-2}{x+2}-\frac{x^2+2}{x^2+2x}=\frac{3}{x}\) ĐKXĐ: \(x\ne0;x\ne-2\)

\(\Leftrightarrow\frac{x\left(x-2\right)}{x\left(x+2\right)}-\frac{x^2+2}{x\left(x+2\right)}=\frac{3\left(x+2\right)}{x\left(x+2\right)}\)

\(\Rightarrow x\left(x-2\right)-\left(x^2+2\right)=3\left(x+2\right)\)

\(\Leftrightarrow x^2-2x-x^2-2=3x+6\)

\(\Leftrightarrow x^2-x^2-2x-3x=2+6\)

\(\Leftrightarrow-5x=8\)

\(\Leftrightarrow x=-\frac{8}{5}\)

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