Câu 1.

a) 5x - 4 = 2x + 11 

<=> 5x - 2x = 11 + 4

<=> 3x = 15 <=> x = 5

Vậy pt có nghiệm x = 5

b) 2x + 3 = 0 <=> x = -3/2

Vậy pt có nghiệm x = -3/2

c) x( 3x - 5 ) = 0

<=> x = 0 hoặc 3x - 5 = 0 

<=> x = 0 hoặc x = 5/3

Vậy pt có tập nghiệm S = { 0 ; 5/3 }

d) 5x - 4 = 2x + 1 

<=> 5x - 2x = 1 + 4

<=> 3x = 5 <=> x = 5/3

Vậy pt có nghiệm x = 5/3

Câu 3.

a) Xét ΔHBA và ΔABC có :

^A = ^H ( = 90⁰ )

^B chung

=> ΔHBA ~ ΔABC ( g.g )

=> HB/AB = AB/BC = HA/AC

=> AB² = HB.BC ( đpcm )

b) Vì ΔABC vuông tại A, áp dụng định lí Pythagoras ta có :

\(BC^2=AB^2+AC^2\)=> \(BC=\sqrt{AB^2+AC^2}=\sqrt{12^2+16^2}=20\left(cm\right)\)

Theo ý a) ta có : HB/AB = AB/BC

<=> HB/12 = 12/20

<=> HB = 12/20 . 12 = 36/5(cm)

c) ΔHBA ~ ΔABC => ^HAB = ^ACB ( hai góc tương ứng )

Xét ΔHAB và ΔHCA có :

^AHB = ^AHC ( = 90⁰ )

^HAB = ^HCA ( cmt ) 

=> ΔHAB ~ ΔHCA ( g.g )

=> HA/HC = AC/Ac = HB/HA

=> AH² = BH.CH ( đpcm )

\(\frac{1}{x^2+2x}+\frac{1}{x^2+6x+8}+\frac{1}{x^2+10x+24}+\frac{1}{x^2+14x+48}=\frac{4}{105}\)

ĐKXĐ : x ≠ 0 ; x ≠ -2 ; x ≠ -4 ; x ≠ -6 ; x ≠ -8

<=> \(\frac{1}{x\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+8\right)}=\frac{4}{105}\)

<=> \(\frac{1}{2}\left[\frac{2}{x\left(x+2\right)}+\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{2}{\left(x+4\right)\left(x+6\right)}+\frac{2}{\left(x+6\right)\left(x+8\right)}\right]=\frac{4}{105}\)

<=> \(\frac{1}{2}\left(\frac{1}{x}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+8}\right)=\frac{4}{105}\)

<=> \(\frac{1}{2}\left(\frac{1}{x}-\frac{1}{x+8}\right)=\frac{4}{105}\)

<=> \(\frac{x+8}{x\left(x+8\right)}-\frac{x}{x\left(x+8\right)}=\frac{8}{105}\)

<=> \(\frac{8}{x\left(x+8\right)}=\frac{8}{105}\)

=> x( x + 8 ) = 105

<=> x² + 8x - 105 = 0

<=> x² - 7x + 15x - 105 = 0

<=> x( x - 7 ) + 15( x - 7 ) = 0

<=> ( x - 7 )( x + 15 ) = 0

<=> x = 7 hoặc x = -15 (tm)

Vậy ...

\(\left(x+3\right)^4+\left(x+5\right)^4=2\)

Đặt \(x+4=a\), phương trình trở thành:

\(\left(a-1\right)^4+\left(a+1\right)^4=2\)

\(\Leftrightarrow a^4-4a^3+6a^2-4a+1+a^4+4a^3+6a^2+4a+1=2\)

\(\Leftrightarrow2a^4+12a^2+2=2\)

\(\Leftrightarrow2a^2\left(a^2+6\right)=0\)

\(\Leftrightarrow a^2\left(a^2+6\right)=0\)

\(\Leftrightarrow\left(x+4\right)^2\left[\left(x+4\right)^2+6\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+4\right)^2=0\\\left(x+4\right)^2+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+4=0\\\left(x+4\right)^2=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-4\\\left(x+4\right)^2=-6\left(vn\right)\end{cases}}\)(vn : vô nghiệm).

Vậy phương trình có nghiệm duy nhất : \(x=-4\)

Đặt t = x + 4

pt <=> ( t - 1 )⁴ + ( t + 1 )⁴ = 2 ( khai triển giống bạn Phạm Thành Đông nhé, mình k làm lại )

<=> 2t⁴ + 12t² + 2 = 2

<=> 2t⁴ + 12t² = 0

<=> t⁴ + 6t² = 0

<=> t²( t² + 6 ) = 0

<=> ( x + 4 )²[ ( x + 4 )² + 6 ] = 0

Vì ( x + 4 )² + 6 ≥ 6 > 0 ∀ x

nên pt <=> ( x + 4 )² = 0 <=> x = -4

Vậy ....

Với \(x\ne1\)ta có 

\(P=\left(\frac{4}{x-1}-\frac{7x+5}{x^3-1}\right):\left(1-\frac{x-4}{x^2+x+1}\right)\)

\(=\left[\frac{4x^2+4x+4-7x-5}{\left(x-1\right)\left(x^2+x+1\right)}\right]:\left(\frac{x^2+x+1-x-4}{x^2+x+1}\right)\)

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

+) 2x² - 5x + 2 < 0 

<=> ( x - 2 )( 2x - 1 ) < 0 

Xét hai trường hợp :

1. \(\hept{\begin{cases}x-2>0\\2x-1< 0\end{cases}}\Rightarrow\hept{\begin{cases}x>2\\x< \frac{1}{2}\end{cases}\left(loai\right)}\)

2. \(\hept{\begin{cases}x-2< 0\\2x-1>0\end{cases}}\Rightarrow\hept{\begin{cases}x< 2\\x>\frac{1}{2}\end{cases}}\Rightarrow\frac{1}{2}< x< 2\)(1)

+) 2x - 3 > 0 <=> x > 3/2 (2)

Từ (1) và (2) => Với 1/2 < x < 3/2 thỏa mãn cả hai bpt trên 

a, \(\frac{x-2}{4}< \frac{x+1}{6}\Leftrightarrow\frac{3x-6}{12}< \frac{2x+2}{12}\)

\(\Rightarrow3x-6< 2x+2\Leftrightarrow-x< 8\Leftrightarrow x>8\)

b, \(2-\frac{x-1}{3}\le\frac{8x+9}{5}\)

\(\Leftrightarrow\frac{7-x}{3}\le\frac{8x+9}{5}\Leftrightarrow\frac{35-5x}{15}\le\frac{24x+27}{15}\)

\(\Rightarrow35-5x\le24x+27\Leftrightarrow8\le29x\Rightarrow x\ge\frac{8}{29}\)

1. \(\frac{x-2}{4}< \frac{x+1}{6}\)

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

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

<=> \(\frac{1}{12}x< \frac{2}{3}\)

<=> x < 8

Vậy ...

2. \(2-\frac{x-1}{3}\le\frac{8x+9}{5}\)

<=> \(2-\frac{1}{3}x+\frac{1}{3}\le\frac{8}{5}x+\frac{9}{5}\)

<=> \(-\frac{1}{3}x-\frac{8}{5}x\le\frac{9}{5}-\frac{7}{3}\)

<=> \(-\frac{29}{15}x\le-\frac{8}{15}\)

<=> \(x\ge\frac{8}{29}\)

Vậy ...

Trả lời:

\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\left(ĐKXĐ:x\ne0;x\ne2\right)\)

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

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

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

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

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{cases}}\)

Vậy S = { -1 }

\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\)ĐK : \(x\ne0;2\)

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

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

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

Vậy tập nghiệm của phương trình là S = { -1 } 

b) \(\frac{x-1}{2x-1}\ge1\)( ĐKXĐ : x ≠ 1/2 )

<=> \(\frac{x-1}{2x-1}-1\ge0\)

<=> \(\frac{x-1}{2x-1}-\frac{2x-1}{2x-1}\ge0\)

<=> \(\frac{x-1-2x+1}{2x-1}\ge0\)

<=> \(\frac{-x}{2x-1}\ge0\)

Xét hai trường hợp :

1. \(\hept{\begin{cases}-x\ge0\\2x-1>0\end{cases}}\Rightarrow loai\)

2. \(\hept{\begin{cases}-x\le0\\2x-1< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge0\\x< \frac{1}{2}\end{cases}}\Rightarrow0\le x< \frac{1}{2}\)

Vậy với \(0\le x< \frac{1}{2}\)thì \(\frac{x-1}{2x-1}\ge1\)

\(\frac{1}{x-8}< \frac{2}{x-6}\)( ĐKXĐ : x ≠ 8 ; x ≠ 6 )

<=> \(\frac{1}{x-8}-\frac{2}{x-6}< 0\)

<=> \(\frac{x-6}{\left(x-8\right)\left(x-6\right)}-\frac{2\left(x-8\right)}{\left(x-8\right)\left(x-6\right)}< 0\)

<=> \(\frac{x-6-2x+16}{\left(x-8\right)\left(x-6\right)}< 0\)

<=> \(\frac{-x+10}{\left(x-8\right)\left(x-6\right)}< 0\)

Xét hai trường hợp :

1.\(\hept{\begin{cases}-x+10>0\\\left(x-8\right)\left(x-6\right)< 0\end{cases}}\Rightarrow6< x< 8\)

2. \(\hept{\begin{cases}-x+10< 0\\\left(x-8\right)\left(x-6\right)>0\end{cases}}\Rightarrow x>10\)

Vậy \(\orbr{\begin{cases}6< x< 8\\x>10\end{cases}}\)thì \(\frac{1}{x-8}< \frac{2}{x-6}\)