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}

3x - 2 < 0

<=> 3x < 2

<=> 3x : 3 < 2 : 3

<=> x < 2/3

Vậy nghiệm của bpt là x < 2/3

3 - 4x \(\ge\)0

<=> -4x \(\ge\)3

<=> -4x : ( -4 )\(\le\)3 : ( -4 )

<=> x \(\le\)-3/4

Vậy nghiệm của bpt là x \(\le\)-3/4

\(\frac{3x}{-5}>0\)

<=> 3x < 0 ( nhân hai vế với -5 và đổi chiều )

<=> 3x : 3 < 0 : 3

<=> x < 0 

Vậy nghiệm của bpt là x < 0

d) Đang tính 

a)

\(\left(x^2-1\right)\left(x^2+4x+3\right)=\left(x-1\right)\left(x+1\right)\left[\left(x+2\right)^2-1\right]=\left(x-1\right)\left(x+1\right)\left(x+1\right)\left(x+3\right)\)

\(\left[\left(x-1\right)\left(x+3\right)\right]\left[\left(x+1\right)\left(x+1\right)\right]=\left(x^2+2x-3\right)\left(x^2+2x+1\right)\)

dặt x^2+2x-1=t(*)

(a) \(\Leftrightarrow\left(t-2\right)\left(t+2\right)=192\) \(\Leftrightarrow t^2-4=192\Rightarrow t^2=196\Rightarrow\left\{\begin{matrix}t=-14\\t=14\end{matrix}\right.\)

Thay t vào (*) => x (tự làm)

a) (x-1)(x+1)(x+1)(x+3)=192. \(\Leftrightarrow\) (x+1)²(x-1)(x+3)=192 \(\Leftrightarrow\) (x²+2x+1) (x²+2x-3)=192 Đặt x²+2x+1=t thì x²+2x-3=t-4 ta có t(t-4)=192 \(\Leftrightarrow\) t²-4t-192=0 \(\Leftrightarrow\) t=-12 hoặc t=16 Với t=-12 thì (x+1)²=-12 ( vô lí ) Với t=16 thì (x+1)²=16 \(\Leftrightarrow\) x=-5 hoặc x=3 b) x\(^5\)+x⁴-2x⁴-2x³+5x³+5x²-2x²-2x+x+1=0 \(\Leftrightarrow\) x⁴(x+1)-2x³(x+1)+5x²(x+1)-2x(x+1)+(x+1)=0 \(\Leftrightarrow\) (x+1)(x⁴-2x³+5x²-2x+1)=0 \(\Leftrightarrow\) x=-1 ( CM x⁴-2x³+5x²-2x+1 vô nghiệm ) c) x⁴-x³-2x³+2x²+2x²-2x-x+1=0 \(\Leftrightarrow\) x³(x-1)-2x²(x-1)+2x(x-1)-(x-1)=0 \(\Leftrightarrow\) (x-1)(x³-2x²+2x-1)=0 \(\Leftrightarrow\) (x-1)(x-1)(x²-x+1)=0 \(\Leftrightarrow\) x-1=0 ( vì x²-x+1=(x-\(\frac{1}{2}\))²+\(\frac{3}{4}\)>0 với mọi x) \(\Leftrightarrow\) x=1

Ở phần b chứng minh vô nghiệm là ( x\(^4\)-2x³+x²)+(3x²-3x+\(\frac{3}{4}\))+\(\frac{5}{4}\)=0 \(\Leftrightarrow\) (x²-x)²+3(x+\(\frac{1}{2}\))²+\(\frac{5}{4}\)=0 ( vô lí)

| 2-4x | = 4x-2

<=> \(\orbr{\begin{cases}\left|2-4x\right|=-2+4x=4x-2\\\left|2-4x\right|=2-4x=4x-2\end{cases}}\)

<=>\(\orbr{\begin{cases}-2+4x=4x-2\\2-4x=4x-2\end{cases}}\)

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

<=>\(\orbr{\begin{cases}0=0\\-8x+4=0\end{cases}}\)

<=> x=\(\frac{-4}{-8}=\frac{1}{2}\)

=> \(S=\left\{\frac{1}{2};\infty\right\}\)

2x-7> 3(x-1)

<=>2x-7>3x-3

<=>2x-3x>-3+7

<=>-x>4

<=>x<4

=>S={x/x<4}

1-2x<4(3x-2)

<=>1-2x<12x-8

<=>-2x-12x<-8-1

<=>-14x<-9

<=>x>\(\frac{9}{14}\)

=>S={\(\frac{9}{14}\)}

-3x+2|-4 -x|> 0

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

<=>\(\orbr{\begin{cases}-2x+6>0\\-8x+2>0\end{cases}}\)

<=>\(\orbr{\begin{cases}-2x>-6\\-8x>-2\end{cases}}\)

<=>\(\orbr{\begin{cases}x< 3\\x< \frac{1}{4}\end{cases}}\)

=>S={x/x<3;x/x<\(\frac{1}{4}\)}

4x-1|x-2|< 0

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

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

<=>\(\orbr{\begin{cases}3x< -1\\3x< 3\end{cases}}\)

<=>\(\orbr{\begin{cases}x< \frac{-1}{3}\\x< 1\end{cases}}\)

=>S={x/x<\(\frac{-1}{3}\);x/x<1}

d) \(PT\Leftrightarrow x\left(2x-7\right)-4\left(x-7\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{7}{2};4\right\}\)

e) \(PT\Leftrightarrow\left(2x-5-x-2\right)\left(2x-5+x+2\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)

Vậy: \(S=\left\{7;1\right\}\)

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

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

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

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

Vậy: \(S=\left\{1;3\right\}\)

\(d,x\left(2x-7\right)-4x+14=0\)

\(x\left(2x-7\right)-2\left(2x-7\right)=0\)

\(\left(x-2\right)\left(2x-7\right)=0\)

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

d: =>(2x-7)(x-2)=0

=>x=7/2 hoặc x=2

e: =>(2x-5-x-2)(2x-5+x+2)=0

=>(x-7)(3x-3)=0

=>x=7 hoặc x=1

f: =>x(x-1)-3(x-1)=0

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

=>x=1 hoặc x=3

\(x^4+x^3+\frac{1}{4}x^2+3x^2-3x+\frac{3}{4}+\frac{3}{4}x^2+\frac{17}{4}=0\)

\(\Leftrightarrow x^2\left(x^2+x+\frac{1}{4}\right)+3\left(x^2-x+\frac{1}{4}\right)+\frac{3}{4}x^2+\frac{17}{4}=0\)

\(\Leftrightarrow x^2\left(x+\frac{1}{2}\right)^2+3\left(x-\frac{1}{2}\right)^2+\frac{3}{4}x^2+\frac{17}{4}=0\)

Phương trình vô nghiệm

a.

ĐKXĐ: \(-1\le x\le1\)

Đặt \(\sqrt{1-x^2}=t\Rightarrow0\le t\le1\)

\(x^2=1-t^2\Rightarrow x^4=t^4-2t^2+1\)

Pt trở thành:

\(729\left(t^4-2t^2+1\right)+8t=36\)

\(\Leftrightarrow729t^4-1458t^2+8t+693=0\)

\(\Leftrightarrow\left(9t^2+2t-9\right)\left(81t^2-18t-77\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}9t^2+2t-9=0\\81t^2-18t-77=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{\sqrt{82}-1}{9}\\t=\dfrac{1+\sqrt{78}}{9}\end{matrix}\right.\)

\(\Rightarrow x=\pm\sqrt{1-t^2}=...\)

b.

ĐKXĐ: ...

\(-3\left(10+4x-x^2\right)-5\sqrt{10+4x-x^2}+42=0\)

Đặt \(\sqrt{10+4x-x^2}=t\ge0\)

\(\Rightarrow-3t^2-5t+42=0\)

\(\Rightarrow\left[{}\begin{matrix}t=3\\t=-\dfrac{14}{3}\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{10+4x-x^2}=3\)

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

\(\Leftrightarrow x=...\)

x + 3x + 4x + 3x + 1 = 0

⇒x + x + 2x + 2x + 2x + 2x + x + 1 = 0

⇒x x + 1 + 2x x + 1 + 2x x + 1 + x + 1 = 0 ⇒ x + 1 x + x + x + x + x + 1 = 0 ⇒ x + 1 x x + 1 + x x + 1 + x + 1 = 0 ⇒ x + 1 x + 1 x + x + 1 = 0 ⇒ x + 1 x + x + 1 = 0 ⇒ x + 1 = 0 vix̀ + x + 1 ≠ 0 ⇒x + 1 = 0 ⇒x = −1 vậy pt có No ......... 3 2x − 3 − 6 x − 3 = 5 4x + 3 − 17 ⇔ 30 10 2x − 3 − 30 5 x − 3 = 30 6 4x + 3 − 30 17.30 ⇔20x − 30 − 5x + 15 = 24x + 18 − 510 ⇔20x − 5x − 24x = 18 − 510 + 30 − 15

⇔− 9x = −477 ⇔x = 53

vậy pt có No........

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

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

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

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

\(\Rightarrow\left(x+1\right)\left[x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\right]=0\)

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

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

\(\Rightarrow\left(x+1\right)^2=0\left(vìx^2+x+1\ne0\right)\)

\(\Rightarrow x+1=0\)

\(\Rightarrow x=-1\)

vậy pt có No .........

\(\frac{2x-3}{3}-\frac{x-3}{6}=\frac{4x+3}{5}-17\)

\(\Leftrightarrow\frac{10\left(2x-3\right)}{30}-\frac{5\left(x-3\right)}{30}=\frac{6\left(4x+3\right)}{30}-\frac{17.30}{30}\)

\(\Leftrightarrow20x-30-5x+15=24x+18-510\)

\(\Leftrightarrow20x-5x-24x=18-510+30-15\)

\(\Leftrightarrow-9x=-477\)

\(\Leftrightarrow x=53\)

vậy pt có No........