Rảnh rỗi thật sự .-.

Bài 1:

a)(4x-3)(3x+2)-(6x+1)(2x-5)+1

=12x²-x-6-12x²+28x+5+1

=27x

b)(3x+4)²+(4x-1)²+(2+5x)(2-5x)

=9x²+24x+16+16x²-8x+1+4-25x²

=16x+21

c)(2x+1)(4x²-2x+1)+(2-3x)(4+6x+9x²)-9

=8x³+1+8-27x³-9

=-19x³

Bài 2:

a)3x(x-4)-x(5+3x)=-34

=>3x²-12x-3x²-5x=-34

=>-17x=-34

=>x=2

Vậy x=2

b)(3x+1)²+(5x-2)²=34(x+2)(x-2)

=>9x²+6x+1+25x²-20x+4=34(x²-4)

=>34x²-14x+5-34x²+136=0

=>-14x+141=0

=>-14x=-141

=>x=\(\frac{141}{14}\)

Vậy x=\(\frac{141}{14}\)

c)x³+3x²+3x+28=0

=>x³-x²+7x+4x²-4x+28=0

=>x(x²-x+7)+4(x²-x+7)=0

=>(x+4)(x²-x+7)=0

\(\Rightarrow\left[\begin{array}{nghiempt}x+4=0\\x^2-x+7=0\left(2\right)\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=-4\\\left(2\right)\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{27}{4}>0\end{array}\right.\)

=>(2) vô nghiệm

Vậy x=-4

bn đòi kq lm sao mk đc tick đây trời thôi t giải hết nhé

a) \(3x\left(2x+1\right)=5\left(2x+1\right)\)

\(3x=5\)

\(x=\frac{5}{3}\)

b) \(\left(3x-8\right)^2=\left(2x-7\right)^2\)

\(3x-8=2x-7\)

\(x=1\)

c) \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2=0\)

\(\left(4x^2-3x-18\right)^2=\left(4x^2+3x\right)^2\)

\(4x^2-3x-18=4x^2+3x\)

\(6x=-18\)

\(x=-3\)

d) Sai đề

e) ko bt

a, (3x - 5)(2x - 1) - (x + 2)(6x - 1) = 0

=> 6x^2 - 3x - 10x + 5 - (6x^2 - x + 12x - 2) = 0

=> 6x^2 - 13x + 5 - 6x^2 - 11x + 2 = 0

=> -24x + 7 = 0 

=> - 24x = -7

=> x = 7/24

b, (3x - 2)(3x + 2) - (3x - 1)^2 = -5

=> 9x^2 - 4 - 9x^2 + 6x - 1 = -5

=> 6x - 5 = -5

=> 6x = 0

=> x = 0

c, x^2 = -6x - 8

=> x^2 + 6x + 8 = 0

=> x^2 + 2.x.3 + 9 - 1 = 0

=> (x + 3)^2 = 1

=> x + 3 = 1 hoặc x + 3 = -1

=> x = -2 hoặc x = -4

a) \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)\left(5x+8\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(3x+1-5x-8\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(-2x-7\right)=0\)

\(TH_1:3x-1=0\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

\(TH_2:-2x-7=0\)

\(\Leftrightarrow x=-\dfrac{7}{2}\)

Vậy pt có tập nghiệm \(S=\left\{\dfrac{1}{3};-\dfrac{7}{2}\right\}\)

b) \(2x^3-5x^2+3x=0\)

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

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

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

\(TH_1:x=0\)

\(TH_2:x-1=0\)

\(\Leftrightarrow x=1\)

\(TH_3:2x-3=0\)

\(\Leftrightarrow x=\dfrac{3}{2}\)

Vậy pt có tập nghiệm \(S=\left\{0;1;\dfrac{3}{2}\right\}\)

c) \(9x^2-16-x\left(3x+4\right)=0\)

\(\Leftrightarrow\left(9x^2-16\right)-x\left(3x+4\right)=0\)

\(\Leftrightarrow\left(3x-4\right)\left(3x+4\right)-x\left(3x+4\right)=0\)

\(\Leftrightarrow\left(3x+4\right)\left(2x-4\right)=0\)

\(TH_1:3x+4=0\)

\(\Leftrightarrow x=-\dfrac{4}{3}\)

\(TH_2:2x-4=0\)

\(\Leftrightarrow x=2\)

Vậy pt có tập nghiệm \(S=\left\{-\dfrac{4}{3};2\right\}\)

d) \(\dfrac{5x+4}{3}-1=\dfrac{3x-2}{4}\)

\(\Leftrightarrow\dfrac{20x+16}{12}-\dfrac{12}{12}=\dfrac{9x-6}{12}\)

\(\Rightarrow20x+16-12=9x-6\)

\(\Leftrightarrow20x-9x=-6-16+12\)

\(\Leftrightarrow11x=-10\)

\(\Leftrightarrow x=-\dfrac{10}{11}\)

Vậy pt có nghiệm duy nhất \(x=-\dfrac{10}{11}\)

a) Ta có: \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)=\left(3x-1\right)\left(5x+8\right)\)

\(\Leftrightarrow3x+1=5x+8\)

\(\Leftrightarrow3x-5x=8-1\)

\(\Leftrightarrow-2x=7\)

\(\Leftrightarrow x=\dfrac{-7}{2}\)

Vậy \(X=\dfrac{-7}{2}\)

b) Ta có: \(2x^3-5x^2+3x=0\)

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

\(\Leftrightarrow x\left[\left(2x^2-2x\right)-\left(3x-3\right)\right]=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\2x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy \(x=1\) hoặc \(x=0\) hoặc \(x=\dfrac{3}{2}\)

c) \(9x^2-16-x\left(3x+4\right)=0\)

\(\Leftrightarrow9x^2-16-3x^2-4x=0\)

\(\Leftrightarrow6x^2-4x-16=0\)

\(\Leftrightarrow2\left(3x^2-2x-8\right)=0\)

\(\Leftrightarrow3x^2-6x+4x-8=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-4}{3}\end{matrix}\right.\)

Vậy \(x=2\) hoặc \(x=\dfrac{-4}{3}\)

d) \(\dfrac{5x+4}{3}-1=\dfrac{3x-2}{4}\)

\(\Leftrightarrow\dfrac{20x+16}{12}-\dfrac{12}{12}=\dfrac{9x-6}{12}\)

\(\Leftrightarrow20x+16-12=9x-6\)

\(\Leftrightarrow20x+16-12-9x+6=0\)

\(\Leftrightarrow11x+10=0\)

\(\Leftrightarrow x=\dfrac{-10}{11}\)

Vậy \(x=\dfrac{-10}{11}\)

a) Ta có: \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)\left(5x+8\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(3x+1-5x-8\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(-2x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\-2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\-2x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{1}{3};-\dfrac{7}{2}\right\}\)

x³ + 3x² + 3x = 7

<=> x³ + 3x² + 3x - 7 = 0

<=> (x - 1)(x² + 4x + 7) = 0

<=> x - 1 = 0 hoặc x² + 4x + 7 khác 0

<=> x - 1 = 0

<=> x = 1

a) ( x² + 3 x + 2 ) . ( x² + 3x+ 3 ) - 2 =0

<=>x⁴ + 3x³ + 3x² + 3x³ + 9x² + 9x + 2x² + 6x + 6 - 2 = 0

<=> x⁴ +  6x³  + 14x² + 15x + 4 = 0

<=> x⁴ + 3x³ + 3x³ + x² + 9x² + 4x² + 3x + 12x + 4 = 0

<=> x² . ( x² +3x + 1 ) + 3x . ( x² +3x + 1 ) + 4. ( x² + 3x + 1 ) = 0

<=> ( x² + 3x + 1 ) . ( x² + 3x + 4 ) = 0

<=> \(\orbr{\begin{cases}x^2+3x+1=0\\x^2+3x+4=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\frac{-3+\sqrt{5}}{2}\\x=\frac{-3-\sqrt{5}}{2}\end{cases}}\)

          \(x\notinℝ\)

<=> \(\orbr{\begin{cases}x=\frac{-3+\sqrt{5}}{2}\\x=\frac{-3-\sqrt{5}}{2}\end{cases}}\)

Nghiệm cuối cùng là : x₁ = \(\frac{-3+\sqrt{5}}{2}\);x₂ = \(\frac{-3-\sqrt{5}}{2}\)

b) ( x + 1 ) . ( x + 2 ) . ( x + 3 ) . ( x + 4 )  - 24 = 0

<=> ( x² + 2x + x + 2 ) . ( x + 3 ) . ( x + 4 ) - 24 = 0

<=> ( x² + 3.x + 2 ) . ( x+3) . ( x + 4 ) -24         = 0

<=> ( x³ + 3.x ² + 3.x² + 9x + 2x + 6 ) . ( x + 4 ) - 24 = 0 

<=> ( x³ + 3x + 2 ) . ( x + 3 ) .( x + 4 ) = 0

<=> ( x³ + 3x² + 3x² + 9x + 2x + 6 ) . ( x + 4) - 24 = 0

<=> ( x³ + 6.x² + 11.x + 6 ) . ( x + 4 ) -24 = 0 

<=> x⁴ + 4.x³ + 6.x³ + 24.x² + 11.x² + 44.x + 6.x + 24 - 24 =0

<=> x⁴ + 10.x³+ 35. x²  + 50.x = 0

<=> x. ( x³ + 10.x² + 35 .x + 50 ) = 0

<=> x. ( x³ + 5.x² +5.x² + 25.x+ 10 + 50 ) = 0

<=> x. [ x² . ( x+5 ) + 5.x. ( x+5 ) + 10.( x + 5 ) ] = 0

<=> x. ( x + 5 ) . ( x² + 5.x + 10 ) = 0

=> \(\hept{\begin{cases}x=0\\x+5=0\\x^2+5.x+10=0\end{cases}}\)

=> \(\hept{\begin{cases}x=0\\x=-5\\x\notinℝ\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-5\\x=0\end{cases}}\)

Nghiệm cuối cùng là : x₁ = -5 ; x₂ = 0

c) x³ + 3.x² + 3x = 7

<=> x³ + 3.x² + 3x - 7 = 0

<=> ( x + 1 )³ - 8          = 0

<=> ( x + 1 )³               = 8

<=> ( x + 1 ) ³               = 2³ 

<=> x + 1                      = 2

<=> x                              =1

Vậy x = 1

a: =>x^2+4x-4x+1=0

=>x^2+1=0

=>Loại

b: =>2x-6+4=2x+2

=>-2=2(loại)

c: =>2(x+3)-2x-1=1

=>6-1=1

=>5=1(loại)

d =>x+3=0

=>x=-3(loại)

e: =>x^2-3x^2+3x-3x-2=0

=>-2x^2-2=0

=>x^2+1=0

=>Loại

a)TH1: \(2x-3>0;3x+2>0\)

\(=>2x-3-3x-2=0\\ =>-x-5=0\\ =>-x=5=>x=-5\)

TH2: \(2x-3< 0;3x+2< 0\)

\(=>-2x+3+3x+2=0\\ =>x+5=0\\ =>x=-5\)

Cả 2 TH ra \(x=-5=>x=-5\)

b)TH1 \(\dfrac{1}{2}x>0\)

\(=>\dfrac{1}{2}x=3-2x\\ =>3-2x-\dfrac{1}{2}x=0\\ =>\dfrac{4}{2}x-\dfrac{1}{2}x=3\\ =>\dfrac{3}{2}x=3\\ =>x=2\)

TH2 \(\dfrac{1}{2}x< 0\)

\(=>-\dfrac{1}{2}x=3-2x\\ =>3-2x+\dfrac{1}{2}x=0\\ =>\dfrac{4}{2}x+\dfrac{1}{2}x=3\\ =>\dfrac{5}{2}x=3\\ =>x=\dfrac{6}{5}\)

\(=>x=2;\dfrac{6}{5}\)

:V lập 2 ý là làm đc á em 

xét 2 trường hợp là đc mà e :v