a, \(5\left(m+3x\right)\left(x+1\right)-4\left(1+2x\right)=80\)

Phương trình nhận \(x=2\)làm nghiệm nên :

\(5\left(m+3.2\right)\left(2+1\right)-4\left(1+2.2\right)=80\)

\(\Leftrightarrow15m+90-20=80\)

\(\Leftrightarrow15m=80+20-90\)

\(\Leftrightarrow15m=10\Leftrightarrow m=1,5\)

....

b, \(3\left(2x+m\right)\left(3x+2\right)-2\left(3x+1\right)^2=43\)

Phương trình nhận \(x=1\)làm nghiệm nên :

\(3\left(2.1+m\right)\left(3.1+2\right)-2\left(3.1+1\right)^2=43\)

\(\Leftrightarrow30+15m-32=43\)

\(\Leftrightarrow15m=43+32-30\)

\(\Leftrightarrow15m=45\Leftrightarrow m=3\)

....

\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)

\(\Leftrightarrow\frac{315-x}{101}+1+\frac{313-x}{103}+1+\frac{311-x}{105}+1+\frac{309-x}{107}+1=0\)

\(\Leftrightarrow\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)

\(\Leftrightarrow\left(416-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)

\(\Leftrightarrow416-x=0\)

\(\Leftrightarrow x=416\)

a) 5(m + 3x)(x + 1) - 4(1 + 2x) = 80

Phương trình có nghiệm x = 2:

5(m + 3.2)(2 + 1) - 4(1 + 2.2) = 80

<=> 5(m + 6).3 - 4.5 = 80

<=> 15(m + 6) - 4.5 = 80

<=> 15(m + 6) - 20 = 80

<=> 15(m + 6) = 80 + 20

<=> 15(m + 6) = 100

<=> m + 6 = 100 : 15

<=> m + 6 = 20/3

<=> m = 20/3 - 6

<=> m = 2/3

b) 3(2x + m)(3x + 2) - 2(3x + 1)² = 43

Phương trình có nghiệm x = 1:

3(2.1 + m)(3.1 + 2) - 2(3.1 + 1)² = 43

<=> 3(2 + m).5 - 2.16 = 43

<=> 15(2 + m) - 32 = 43

<=> 15(2 + m) = 43 + 32

<=> 15(2 + m) = 75

<=> 2 + m = 75 : 15

<=> 2 + m = 5

<=> m = 5 - 2

<=> m = 3

Vì dài quá nên mình chỉ có thể trả lời được mấy câu thôi

Bài 1:

27x³ - 8 : (6x + 9x² +4)

= (3x - 2) (9x² + 6x + 4) : (9x² + 6x + 4)

= 3x - 2

Bài 3:

a, 81x⁴ + 4 = (9x²)² + 36x² + 4 - 36x²

= (9x² + 2)² - (6x)²

= (9x² + 6x + 2)(9x² - 6x + 2)

b, x² + 8x + 15 = x² + 3x + 5x + 15

= x(x + 3) + 5(x + 3)

= (x + 3)(x + 5)

c, x² - x - 12 = x² + 3x - 4x - 12

= x(x + 3) - 4(x + 3)

= (x + 3) (x - 4)

Câu 1:

(27x³ - 8) : (6x + 9x² + 4)

= (3x - 2)(9x² + 6x + 4) : (6x + 9x² + 4)

= 3x - 2

Câu 2:

a) (3x - 5)(2x+ 11) - (2x + 3)(3x + 7)

= 6x² + 33x - 10x - 55 - 6x² - 14x - 9x - 21

= -76

⇒ đccm

b) (2x + 3)(4x² - 6x + 9) - 2(4x³ - 1)

= 8x³ + 27 - 8x³ + 2

= 29

⇒ đccm

Câu 3:

a) 81x⁴ + 4

= (9x²)² + 2²

= (9x² + 2)² - (6x)²

= (9x² - 6x + 2)(9x² + 6x + 2)

b) x² + 8x + 15

= x² + 3x + 5x + 15

= x(x + 3) + 5(x + 3)

= (x + 3)(x + 5)

c) x² - x - 12

= x² - 4x + 3x - 12

= x(x - 4) + 3(x - 4)

= (x - 4)(x + 3)

Bài 2:

a, \(A=3x\left(2x-5y\right)+\left(3x-y\right)\left(-2x\right)-\dfrac{1}{2}\left(2-26xy\right)\)

\(=6x^2-15xy-6x^2+2xy-1+13xy\)

\(=-1\)

\(\Rightarrowđpcm\)

b, \(B=\left(2x-3\right)\left(4x+1\right)-4\left(x-1\right)\left(2x-1\right)-2x+5\)

\(=8x^2+2x-12x-3-4\left(2x^2-x-2x+1\right)-2x+5\)

\(=8x^2-10x+2-8x^2+4x+8x-4-2x\)

\(=2-4=-2\)

\(\Rightarrowđpcm\)

Bài 1:

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

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

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

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

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

\(\Leftrightarrow x+2=0\) hoặc \(x+4=0\)

\(\Leftrightarrow x=-2\) hoặc \(x=-4\)

b) \(2x^3+\dfrac{3}{2}x^2=0\)

\(\Leftrightarrow x^2\left(2x+\dfrac{3}{2}\right)=0\)

\(\Leftrightarrow x^2=0\) hoặc \(2x+\dfrac{3}{2}=0\)

\(\Leftrightarrow x=0\) hoặc \(x=-\dfrac{3}{4}\)

bài 1

a) (x+2)²-x²+4=0

\(\Leftrightarrow\)x²+4x+4-x²+4=0

\(\Leftrightarrow\)4x+8=0

\(\Leftrightarrow\) 4(x+2)=0

=>x+2=0

\(\Leftrightarrow\)x=-2

vậy x=-2

b) \(2x^3+\dfrac{3}{2}x^2=0\)

\(\Leftrightarrow x^2\left(2x+\dfrac{3}{2}\right)=0\)

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

vậy x=0 hoặc x=-\(\dfrac{3}{4}\)

bài 1:

b,\(\dfrac{x+2}{x}=\dfrac{x^2+5x+4}{x^2+2x}+\dfrac{x}{x+2}\)(ĐKXĐ:x ≠0,x≠-2)

<=>\(\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x^2+5x+4}{x\left(x+2\right)}+\dfrac{x^2}{x\left(x+2\right)}\)

=>\(x^2+4x+4=x^2+5x+4+x^2\)

<=>\(x^2-x^2-x^2+4x-5x+4-4=0\)

<=>\(-x^2-x=0< =>-x\left(x+1\right)=0< =>\left[{}\begin{matrix}x=0\left(loại\right)\\x+1=0< =>x=-1\left(nhận\right)\end{matrix}\right.\)

vậy...............

d,\(\left(x+3\right)^2-25=0< =>\left(x+3-5\right)\left(x+3+5\right)=0< =>\left(x-2\right)\left(x+8\right)=0< =>\left[{}\begin{matrix}x-2=0\\x+8=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)

vậy............

bài 3:

g,\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-x-2}\)(ĐKXĐ:x khác -1,x khác 2)

<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-2x+x-2}\)

<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x\left(x-2\right)+\left(x-2\right)}\)

<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)

<=>\(\dfrac{4\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{2\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)

=>\(4x-8-2x-2=x+3\)

<=>\(x=13\)

vậy..............

mấy ý khác bạn làm tương tụ nhé

chúc bạn học tốt ^ ^

a) \(\left(2x+1\right)^2-\left(x+2\right)^2>0\)

\(\Leftrightarrow\left(2x+1-x-2\right)\left(2x+1+x+2\right)>0\)

\(\Leftrightarrow\left(x-1\right)\left(3x+3\right)>0\)

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

Vậy tập nghiệm của bất phương trình là x > 1 hoặc x < -1

b) Sửa lại rồi làm câu b nèk\(\dfrac{5x-3x}{5}+\dfrac{3x+1}{4}>\dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\)

\(\Leftrightarrow4\left(5x-3x\right)+5\left(3x+1\right)>10\left(x+2x\right)-30\)\(\Leftrightarrow20x-12x+15x+5>10x+20x-30\)\(\Leftrightarrow20x-12x+15x-10x-20x>-30-5\)\(\Leftrightarrow-7x>-35\)

\(\Leftrightarrow x< 5\)

c) \(\dfrac{-1}{2x+3}< 0\)

dễ nhé mình học bài hóa mai kt 15 phút nên ko có time để giúp

a) \(A = \frac{2x^2 - 16x+43}{x^2-8x+22}\) = \(\frac{2(x^2-8x+22)-1}{x^2-8x+22}\) = \(2 - \frac{1}{x^2-8x+22}\)

Ta có : \(x^2-8x+22 \) = \(x^2-8x+16+6 = ( x-4)^2 +6 \)

Vì \((x-4)^2 \ge 0 \) với \( \forall x\in R\) Nên \(( x-4)^2 +6 \ge 6 \)

\(\Rightarrow \) \(x^2-8x+22 \) \( \ge 6\)\(\Rightarrow \) \(\frac{1}{x^2-8x+22} \) \(\le \frac{1}{6}\) \(\Rightarrow \) - \(\frac{1}{x^2-8x+22} \) \(\ge - \frac{1}{6}\)

\(\Rightarrow \) A = \(2 - \frac{1}{x^2-8x+22}\) \( \ge 2-\frac{1}{6}\) = \(\frac{11}{6}\) Dấu "=" xảy ra khi và chỉ khi x=4

Vậy GTNN của A = \(\frac{11}{6}\) khi và chỉ khi x=4

cần giúp ko

cách khác

Chứng minh điều ngược lại

Câu 2: phần (b)

\(A=\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=\dfrac{\left(a+b\right)^2}{ab}\)

Để tồn tại A => a,b khác 0

\(\left(a+b\right)^2\ge0\forall a,b\in R\)

\(ab>0\) có một chút hy vọng \(A\ge4\)

ab< 0 tắt hét hy vọng \(A\ge4\)

Kết luận chẳng hy vọng gì

cần face với thực tế Quý Minh mình đã giải thích cho bạn thế nào là face đó

a) \(15x-3\left(3x-2\right)=45-5\left(2x-5\right)\)

\(\Leftrightarrow15x-9x+6=45-10x+25\)

\(\Leftrightarrow15x-9x+10x=45+25-6\)

\(\Leftrightarrow16x=64\)

\(\Leftrightarrow x=4\)

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

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

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

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

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

c) \(\dfrac{1}{x-4}+\dfrac{x+2}{x+4}=\dfrac{5x-4}{x^2-16}\)

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

\(\Leftrightarrow x+4+x^2-4x+2x-8=5x-4\)

\(\Leftrightarrow x^2+x-4x+2x-5x=-4+8-4\)

\(\Leftrightarrow x^2-6x=0\)

\(\Leftrightarrow x\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-6=0\Leftrightarrow x=6\end{matrix}\right.\)

a) 15x - 3(3x - 2) = 45 - 5(2x - 5)

\(\Leftrightarrow\) 15x - 9x + 6 = 45 - 10x + 25

\(\Leftrightarrow\) 6x + 10x = 70 - 6

\(\Leftrightarrow\) 16x = 64

\(\Leftrightarrow\) x = 4

Vậy.......................

b) x² - 9 + 4(x - 3) = 0

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

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

\(\Leftrightarrow\) (x - 3)(x + 7) = 0

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

Vậy........................

c) \(\dfrac{1}{x-4}+\dfrac{x+2}{x+4}=\dfrac{5x-4}{x^2-16}\)

\(\Leftrightarrow\) \(\dfrac{1}{x-4}+\dfrac{x+2}{x+4}=\dfrac{5x-4}{\left(x-4\right)\left(x+4\right)}\) (đk: x\(\ne\pm\)4)

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

\(\Leftrightarrow\) x + 4 + x² - 4x + 2x - 8 = 5x - 4

\(\Leftrightarrow\) x² - x - 5x - 4 + 4 = 0

\(\Leftrightarrow\) x² - 6x = 0

\(\Leftrightarrow\) x(x - 6) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tmđk\right)\\x=6\left(tmđk\right)\end{matrix}\right.\)

Vậy...............