a) x=3/2

b)x=-1

c) x=5

d) x= 5/2

/5x-4/=/x+2/

\(\orbr{\begin{cases}5x-4=x+2\\5x-4=-x+2\end{cases}}suyra\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{2}\end{cases}}\)

vậy x=3/2 hoặc x=1/2

Ta có : \(\left|5x-4\right|=\left|x+2\right|\)

\(\Leftrightarrow\orbr{\begin{cases}5x-4=x+2\\5x-4=-x-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x-x=2+4\\5x+x=-2+4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}4x=6\\6x=2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{3}\end{cases}}\)

b) \(\left|2x-3\right|-\left|3x+2\right|=0\)

\(\Rightarrow\orbr{\begin{cases}2x-3=3x+2\\2x-3=-3x-2\end{cases}\Rightarrow\orbr{\begin{cases}2x-3x=2+3\\2x+3x=-2+3\end{cases}\Rightarrow}\orbr{\begin{cases}-x=5\\5x=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=-5\\x=\frac{1}{5}\end{cases}}}\)

c)/2+3x/=/4x-3/

\(\Rightarrow\orbr{\begin{cases}2+3x=4x-3\\2+3x=-\left(4x-3\right)\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x-4x=-3-2\\3x+4x=3-2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}-x=-5\\7x=1\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=\frac{1}{7}\end{cases}}}\)

d)/7x+1/-/5x+6|=0

\(\Rightarrow\left|7x+1\right|=\left|5x+6\right|\)

\(\Rightarrow\orbr{\begin{cases}7x+1=5x+6\\7x+1=-\left(5x+6\right)\end{cases}\Rightarrow\orbr{\begin{cases}7x-5x=6-1\\7x+1=-5x-6\end{cases}\Rightarrow}\orbr{\begin{cases}2x=5\\7x+5x=-6-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{7}{12}\end{cases}}}\)

\(6x^2+x-15=0\Leftrightarrow6x^2+10x-9x-15=0\)

\(\Leftrightarrow2x\left(3x+5\right)-3\left(3x+5\right)=0\Leftrightarrow\left(2x-3\right)\left(3x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=0\\3x+5=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{5}{3}\end{cases}}\)

Tập nghiệm của phương trình là \(S=\left\{\frac{3}{2};-\frac{5}{3}\right\}\)

a) 4(x+2) - 7(2x - 1) + 9(3x - 4)=30

⇔4x+8 - 14x + 7 + 27x - 36 = 30

⇔ 17x = 51

⇔ x = 3

b) 2(5x - 8) - 3(4x - 5) = 4(3x - 4) + 11

⇔ 10x - 16 - 12x + 15 = 12x - 16 + 11

⇔ -14x = -4

⇔ x= \(\frac{2}{7}\)

c) 5x(1 - 2x) - 3x(x + 18) = 0

⇔ 5x - 10x\(^2\) - 3x\(^2\) -54x =0

⇔ -13x\(^2\) -49 x = 0

⇔ -x ( 13x + 49 ) =0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\13x+49=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{-49}{13}\end{matrix}\right.\)

d) 5x - 3{4x - 2[4x - 3(5x - 2)]} = 182

⇔ 5x - 3[ 4x - 2( 4x - 15x + 6 ) ]= 182

⇔5x - 3 ( 4x - 8x + 30x - 12 ) = 182

⇔ 5x - 3 ( 26x - 12 ) = 182

⇔ 5x - 78x + 36 = 182

⇔ - 73x = 146

⇔ x = -2

a) <=> \(2x^2-8x+3x-12+x^2-7x+10=3x^2-5x-12x+20\)

<=> \(2x^2-8x+3x-12+x^2-7x+10-3x^2+5x+12x-20=0\)

<=> \(5x-22=0\)

<=> \(5x=22\)

<=> \(x=\frac{22}{5}\)

b) <=> \(24x^2-9x+16x-6-4x^2-7x-16x-28=10x^2+5x-2x-1\)

<=> \(24x^2-9x+16x-6-4x^2-7x-16x-28-10x^2-5x+2x+1=0\)

<=> \(10x^2-19x-33=0\)

<=> \(10x^2-30x+11x-33=0\)

<=> \(10x\left(x-3\right)+11\left(x-3\right)=0\)

<=> \(\left(x-3\right)\left(10x+11\right)=0\)

<=> \(x=3;x=-\frac{11}{10}\)

2x² - ( m + 4 )x + m = 0

Δ = b² - 4ac = ( m + 4 )² - 8m = m² + 8m + 16 - 8m = m² + 16

Vì m² + 16 ≥ 16 > 0 ∀ m => Δ ≥ 16 > 0

Vậy phương trình luôn có nghiệm ( đpcm )

Bài làm

G = 2x² - 3x + 1

G = 2x² - 2x - x + 1

G = -( 2x² + 2x ) - ( x + 1 )

G = -2x( x + 1 ) - ( x + 1 )

G = ( x + 1 )( -2x - 1 )

# Học tốt #

Bài làm

H = -x² + 5x - 4

H = -x² + 4x + x - 4

H = -( x² - 4x ) + ( x - 4 )

H = -x( x - 4 ) + ( x + 4 )

H = ( x - 4 )( -x + 1 )

# Học tốt #

Bài làm

I = x² + 4x +  3

I = x² + 3x + x + 3

I = ( x² + 3x ) + ( x + 3 )

I = x( x + 3 ) + ( x + 3 )

I = ( x + 3 )( x + 1 )

# Học tốt #