a) 2x² - 3x - 2 = 0.

<=> (2x + 1)(x - 2) = 0

<=> 2x + 1 = 0 hoặc x - 2 = 0

<=> x = -1/2 hoặc x = 2

b) 3x² - 7x - 10 = 0.

<=> (x + 1)(3x - 10) = 0

<=> x = -1 hoặc x = 10/3

c) 2x² - 5x + 3 = 0.

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

<=> x = 1 hoặc x = 3/2

a) x = 2 7                         b) x = 2.

c) x = 2                          d) x = 1.

b) \(\Leftrightarrow3x^3+12x-2x^2-8=0\\ \Leftrightarrow\left(3x^3-2x^2\right)+\left(12x-8\right)=0\\ \Leftrightarrow x^2\left(3x-2\right)+4\left(3x-2\right)=0\\ \Leftrightarrow\left(x^2+4\right)\left(3x-2\right)=0\)

Vì \(x^2+4>0\Rightarrow3x-2=0\Rightarrow x=\dfrac{2}{3}\)

c) \(x^2+5x=0\\ \Leftrightarrow x\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

d) \(\Leftrightarrow x^3-27+x\left(4-x^2\right)=36\\ \Leftrightarrow x^3+4x-x^3=63\\ \Leftrightarrow4x=63\\ \Leftrightarrow x=\dfrac{63}{4}\)

b) 3x(x\(^3\) +12x-2x\(^2\)-8=0

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

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

\(\Leftrightarrow\left[{}\begin{matrix}X^2+4=0\\3X-2=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x\in Z\\X=\dfrac{2}{3}\end{matrix}\right.\)
 

a) x\(^2\)+5x=0

x(x+5)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
 

c)(x-3)(x\(^2\)+3x+9)+x(x+2)(2-x)=36

x\(^3\)-27+x(x+2)(2-x)=36

4x-27=36

4x=36+27

4x=63

x=\(\dfrac{63}{4}\)

a) \(4x^2+12x+1=\left(4x^2+12x+9\right)-8=\left(2x+3\right)^2-8\ge-8\)

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

b) \(4x^2-3x+10=\left(4x^2-3x+\dfrac{9}{16}\right)+\dfrac{151}{16}=\left(2x-\dfrac{3}{4}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\)

\(ĐTXR\Leftrightarrow x=\dfrac{3}{8}\)

c) \(2x^2+5x+10=\left(2x^2+5x+\dfrac{25}{8}\right)+\dfrac{55}{8}=\left(\sqrt{2}x+\dfrac{5\sqrt{2}}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\)

\(ĐTXR\Leftrightarrow x=-\dfrac{5}{4}\)

d) \(x-x^2+2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{9}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\)

\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)

e) \(2x-2x^2=-2\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{2}=-2\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\le\dfrac{1}{2}\)

\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)

f) \(4x^2+2y^2+4xy+4y+5=\left(4x^2+4xy+y^2\right)+\left(y^2+4y+4\right)+1=\left(2x+y\right)^2+\left(y+2\right)^2+1\ge1\)

\(ĐTXR\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

a: Ta có: \(4x^2+12x+1\)

\(=4x^2+12x+9-8\)

\(=\left(2x+3\right)^2-8\ge-8\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{2}\)

b: Ta có: \(4x^2-3x+10\)

\(=4\left(x^2-\dfrac{3}{4}x+\dfrac{5}{2}\right)\)

\(=4\left(x^2-2\cdot x\cdot\dfrac{3}{8}+\dfrac{9}{64}+\dfrac{151}{64}\right)\)

\(=4\left(x-\dfrac{3}{8}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{8}\)

c: Ta có: \(2x^2+5x+10\)

\(=2\left(x^2+\dfrac{5}{2}x+5\right)\)

\(=2\left(x^2+2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}+\dfrac{55}{16}\right)\)

\(=2\left(x+\dfrac{5}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{5}{4}\)

kh hiểu bn ơi

`4x=2+xx+1x<=>4x=2+3x<=>4x-3x=2<=>1x=2<=>x=2`

bạn tách từng bài ra bn

tl câu hỏi trên cho mik ik

giải phương trình làm dell gì, lớp 8 mà

c. Ta có h(x) = 0 ⇒ 5x + 1 = 0 ⇒ x = -1/5

Vậy nghiệm của đa thức h(x) là x = -1/5 (1 điểm)

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

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

\(\left(x-1\right)\left(2x+11\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\2x+11=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\2x=-11\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=-\frac{11}{2}\end{cases}}}\)

\(b,3x\left(25x+15\right)-35\left(5x+3\right)=0\)

\(15x\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\left(5x+3\right).5\left(3x-7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x+3=0\\5\left(3x-7\right)=0\end{cases}\Rightarrow\orbr{\begin{cases}5x=-3\\3x-7=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\x=\frac{7}{3}\end{cases}}}\)

\(c,\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)

\(\left(3x-2\right)\left(2-5x\right)+\left(3x-2\right)\left(x+11\right)=0\)

\(\left(3x-2\right)\left(2-5x+x+11\right)=0\)

\(\left(3x-2\right)\left(13-4x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-2=0\\13-4x=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=2\\4x=13\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{13}{4}\end{cases}}}\)

còn đâu tự lm lười :_# 

2 x 2 - 5 x + 3 x 2 + 5 x + 4 = 2 x 4 + 10 x 3 + 8 x 2 - 5 x 3 - 25 x 2 - 20 x + 3 x 2 + 15 x + 12 = 2 x 4 + 5 x 3 - 14 x 2 - 5 x + 12

x 2 + 3 x - 4 2 x 2 - x - 3 = 2 x 4 - x 3 - 3 x 2 + 6 x 3 - 3 x 2 - 9 x - 8 x 2 + 4 x + 12 = 2 x 4 + 5 x 3 - 4 x 2 - 5 x + 12

Ta có: 2 x 2 - 5 x + 3 x 2 + 5 x + 4 = x 2 + 3 x - 4 2 x 2 - x - 3

Vậy đẳng thức đúng.