Cs vẻ bn đx qw rảnh rỗi òi -,-

4x² + 4x + 1 = 0

( 2x + 1 )² = 0

=> 2x + 1 = 0

=> x = -1/2

Hok tốt

Bài làm:

a) \(4x^2-7x+3=0\)

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

\(\Leftrightarrow4x\left(x-1\right)-3\left(x-1\right)=0\)

\(\Leftrightarrow\left(4x-3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x-3=0\\x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{3}{4}\\x=1\end{cases}}\)

b) \(\left(4x^2-4\right)\left(x^2-x\right)=0\)

\(\Leftrightarrow4x\left(x-1\right)^2\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)(Do viết PT lỗi nên bạn tự giải nha)

c) \(6x^2-4x-2=0\)

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

\(\Leftrightarrow6x\left(x-1\right)+2\left(x-1\right)=0\)

\(\Leftrightarrow2\left(3x+1\right)\left(x-1\right)=0\)

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

Sa

a) \(4x^2-7x+3=0\)

Dễ dàng nhận thấy a + b + c = 4 + ( -7 ) + 3 = 0 

Vậy nên phương trình đã cho có hai nghiệm phân biệt

 \(\hept{\begin{cases}x_1=1\\x_2=\frac{c}{a}=\frac{3}{4}\end{cases}}\)

Vậy \(S=\left\{1;\frac{3}{4}\right\}\)

b) \(\left(4x^2-4\right)\left(x^2-x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x^2-4=0\\x^2-x=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}4\left(x^2-1\right)=0\\x\left(x-1\right)=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^2-1=0\Leftrightarrow x^2=1\Leftrightarrow x=\pm1\\x-1=0\Leftrightarrow x=1\\x=0\end{cases}}\)( chỗ này bạn thay bằng dấu hoặc nhé )

Vậy \(S=\left\{0;\pm1\right\}\)

c) \(6x^2-4x-2=0\)

Dễ dàng nhận thấy a + b + c = 6 + ( -4 ) + ( -2 ) = 0

Vậy nên phương trình đã cho có hai nghiệm phân biệt :

\(\hept{\begin{cases}x_1=1\\x_2=\frac{c}{a}=\frac{-2}{6}=-\frac{1}{3}\end{cases}}\)

Vậy \(S=\left\{1;-\frac{1}{3}\right\}\)

Câu b) bị lỗi hiển thị :v 

\(4\left(x^2-4\right)\left(x^2-x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x^2-4=0\\x^2-x=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}4\left(x^2-1\right)=0\\x\left(x-1\right)=0\end{cases}}\)

<=> x² - 1 = 0 hoặc x = 0 hoặc x - 1 = 0

<=> x = \(\pm\)1 hoặc x = 0

Vậy \(S=\left\{0;\pm1\right\}\)

a)\(2x^4-6x^3+x^2+6x-3=0\)

\(\Leftrightarrow2x^4-6x^3+3x^2-2x^2+6x-3=0\)

\(\Leftrightarrow x^2\left(2x^2-6x+3\right)-\left(2x^2-6x+3\right)=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(2x^2-6x+3\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(2x^2-6x+3\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+1=0\\2x^2-6x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\\Delta_{2x^2-6x+3}=\left(-6\right)^2-4\left(2.3\right)=12\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\x_{1,2}=\frac{6\pm\sqrt{12}}{4}\end{array}\right.\)

b)\(x^3+9x^2+26x+24=0\)

\(\Leftrightarrow x^3+5x^2+6x+4x^2+20x+24=0\)

\(\Leftrightarrow x\left(x^2+5x+6\right)+4\left(x^2+5x+6\right)=0\)

\(\Leftrightarrow\left(x^2+5x+6\right)\left(x+4\right)=0\)

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

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

5.

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\frac{5}{6}\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)

\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)

\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)

\(\Leftrightarrow\frac{1}{3}sin^22x=\frac{1}{6}\)

\(\Leftrightarrow sin^22x=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\frac{\sqrt{2}}{2}\\sin2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=\frac{3\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\\x=\frac{5\pi}{8}+k\pi\end{matrix}\right.\)

6.

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+\frac{1}{2}sinx.cosx=0\)

\(\Leftrightarrow1-3sin^2x.cos^2x+\frac{1}{2}sinx.cosx=0\)

\(\Leftrightarrow1-\frac{3}{4}sin^22x+\frac{1}{4}sin2x=0\)

\(\Leftrightarrow-3sin^22x+sin2x+4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-1\\sin2x=\frac{4}{3}>1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow2x=-\frac{\pi}{2}+k2\pi\)

\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)

1.

\(\Rightarrow4cos^3x.cos3x+4sin^3x.sin3x=\sqrt{2}\)

\(\Leftrightarrow\left(3cosx+cos3x\right)cos3x+\left(3sinx-sin3x\right)sin3x=\sqrt{2}\)

\(\Leftrightarrow3\left(cos3x.cosx+sin3x.sinx\right)+cos^23x-sin^23x=\sqrt{2}\)

\(\Leftrightarrow3cos2x+cos6x=\sqrt{2}\)

\(\Leftrightarrow3cos2x+4cos^32x-3cos2x=\sqrt{2}\)

\(\Leftrightarrow4cos^32x=\sqrt{2}\)

\(\Leftrightarrow cos2x=\frac{\sqrt{2}}{2}\)

\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{4}+k2\pi\\2x=-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\end{matrix}\right.\)

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

3x(x-2)-(x-2)=0

(3x-1)(x-2)=0

=>\(\hept{\begin{cases}3x-1=0\\x-2=0\end{cases}}\)

=> \(\hept{\begin{cases}3x=1\\x=2\end{cases}}\)

=>\(\hept{\begin{cases}x=\frac{1}{3}\\x=2\end{cases}}\)

vậy x thuộc (1/3;2)

b. 4x(x-3)-2x+6=0

4x(x-3) -2(x-3)=0

(4x-2)(x-3)

=>*4x-2=0

4x=2

x=1/2

*x-3=0

x=3

vậy x thuộc (1/2;3)

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

=>(x-2)(3x-1)=0

=>x-2=0 hoặc 3x-1=0

*x-2=0=>x=2

*3x-1=0=>x=\(\frac{1}{3}\)

Vậy x=2;x=\(\frac{1}{3}\)

b. 4x(x-3)-2x+6=0

=>(x-3)(4x-2)=0

=>(x-3)=0 hoặc 4x-2=0

*x-3=0=>x=3

*4x-2=0=>\(\frac{1}{2}\)

Vậy x=3; x=\(\frac{1}{2}\)

câu c, d,e thì chút nx mk gửi cho nha

Giúp mình với

a)4(2x+1)\(^2\)+(4x+2)(2-6x)+(3x-1)\(^2\)=0

=>\(4\left(2x+1+3x-1\right)^2=0\)

=>\(4.\left(5x\right)^2=0\)

=>\(4.25x^2=0\)

=>\(100x^2=0\)

=>\(x^2=0\)

=>x=0

Vậy x =0

câu b đề sai thì phải!

a) 3x² - 4x + 1 = 0

a = 3; b = -4; c = 1

∆ = b² - 4ac = (-4)² - 4.3.1 = 4 > 0

Phương trình có hai nghiệm phân biệt:

x₁ = (-b + √∆)/2a = [-(-4) + 2]/(2.3) = 1

x₂ = (-b - √∆)/2a = [-(-4) - 2]/(2.3) = 1/3

Vậy S = {1/3; 1}

b) -4x² + 4x + 1 = 0

a = -4; b = 4; c = 1

∆ = b² - 4ac = 4² - 4.(-4).1 = 32 > 0

Phương trình có hai nghiệm phân biệt:

x₁ = (-b + √∆)/2a = (-4 + 4√2)/[2.(-4)] = (1 - √2)/2

x₂ = (-b - √∆)/2a = (-4 - 4√2)/[2.(-4)] = (1 + √2)/2

Vậy S = {(1 - √2)/2; (1 + √2)/2}

d) x² - 8x + 2 = 0

a = 1; b = -√8; c = 2

∆ = b² - 4ac = 8 - 8 = 0

Phương trình có nghiệm kép:

x₁ = x₂ = -b/2a = √8/2 = √2

Vậy S = {√2}

e) x² - 6x + 5 = 0

a = 1; b = -6; c = 5

∆ = b² - 4ac = 36 - 20 = 16 > 0

Phương trình có hai nghiệm phân biệt:

x₁ = (-b + √∆)/2a = (6 + 4)/2 = 5

x₂ = (-b - √∆)/2a = (6 - 4)/2 = 1

Vậy S = {1; 5}

ai giúp em với ạ 😥