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15 tháng 11 2017

a, \(D=\dfrac{2x^2+9}{x^2+4}=\dfrac{2x^2+8+1}{x^2+4}=\dfrac{2\left(x^2+4\right)+1}{x^2+4}=2+\dfrac{1}{x^2+4}\)

Suy ra \(D\) lớn nhất \(\Leftrightarrow\dfrac{1}{x^2+4}\) lớn nhất \(\Leftrightarrow x^2+4\) nhỏ nhất

Ta có: \(x^2\ge0\left(\forall x\right)\)\(\Rightarrow x^2+4\ge4\left(\forall x\right)\)

Từ đó ta dễ dàng tìm ra được GTNN của \(x^2+4=4\Leftrightarrow x^2=0\Leftrightarrow x=0\)

Vậy \(MaxD=\) \(2+\dfrac{1}{4}=\dfrac{9}{4}\) \(\Leftrightarrow x=0\)

b,\(Q=\dfrac{5x^2+10x+42}{x^2+2x+7}=\dfrac{5\left(x^2+2x+7\right)+5}{x^2+2x+7}=5+\dfrac{5}{x^2+2x+7}\)

Tương tự câu a, \(Q\) lớn nhất \(\Leftrightarrow x^2+2x+7\) nhỏ nhất

\(x^2+2x+7=x^2+2x+1+6=\left(x+1\right)^2+6\ge6\left(\forall x\right)\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)

Vậy \(MaxQ=5+\dfrac{5}{x^2+2x+7}=5+\dfrac{5}{6}=\dfrac{35}{6}\Leftrightarrow x=-1\)

5: \(\Leftrightarrow9\left(x^2-5x-4\right)=36\left(x+1\right)+8\left(x^2-10x\right)\)

\(\Leftrightarrow9x^2-45x-36-36x-36-8x^2+80x=0\)

\(\Leftrightarrow x^2-x-72=0\)

=>(x-9)(x+8)=0

=>x=9 hoặc x=-8

6: \(\Leftrightarrow x^2-9=9x-x^2-9+x\)

\(\Leftrightarrow2x^2-10x=0\)

=>2x(x-5)=0

=>x=0 hoặc x=5

23 tháng 1 2022

5, <=> 9x^2 - 45x - 36 = 36x + 36 + 8x^2 - 80x 

<=> x^2 - x - 72 = 0 <=> x = 9 ; x = -8 

6, <=> x^2 - 9 = 9x - x^2 - 9 + x = 10x - x^2 - 9 

<=> 2x^2 - 10x = 0 <=> x = 0 ; x = 5 

7, <=> (x-1)^2 = (3x+3)^2 

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

<=> (-2x-4)(4x+2) = 0 <=> x = -2;x=-1/2

8, = (x^2-10x-15)(x^2-10x+25)

29 tháng 1 2022

1.

<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)

2.

<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

3.

<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)

4.

<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

5. 

<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)

6,7. ko đủ điều kiện tìm

29 tháng 1 2022

Oki pạn cảm ơn

 

6 tháng 6 2019

a,\(x-\frac{5x+2}{6}=\frac{7-3x}{4}\)

=> \(\frac{12x}{12}-\frac{\left(5x+2\right)2}{12}=\frac{\left(7-3x\right)3}{12}\)

=>\(\frac{12x-10x-4}{12}=\frac{21-9x}{12}\)

=>(khử mẫu)

=>\(12x-10x-4=21-9x\)

=>11x=25

=>x=25/11

b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)

=>30x+9=36+32x+24

=>32x+60=30x+9

=>2x=-51

=>x=-51/2

c: \(\Leftrightarrow2x-3\left(2x+1\right)=x+6x\)

=>7x=2x-6x-3

=>7x=-4x-3

=>11x=-3

=>x=-3/11

d: \(\Leftrightarrow4\left(x+2\right)-6x=3\left(1-2x+1\right)\)

=>4x+8-6x=3(-2x+2)

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

=>4x+2=0

=>x=-1/2

28 tháng 6 2017

Phép trừ các phân thức đại số

16 tháng 1 2019

a)\(x-\dfrac{5x+2}{6}=\dfrac{7-3x}{4}\)

\(\Leftrightarrow\dfrac{12x-10x-4}{12}=\dfrac{21-9x}{12}\)

\(\Leftrightarrow2x-4=21-9x\)

\(\Leftrightarrow2x-4-21+9x=0\)

\(\Leftrightarrow11x-25=0\)

\(\Leftrightarrow x=\dfrac{25}{11}\)

b)\(\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\)

\(\Leftrightarrow\dfrac{30x+9}{36}=\dfrac{36+24+32x}{36}\)

\(\Leftrightarrow30x+9=60+32x\)

\(\Leftrightarrow30x+9-60-32x=0\)

\(\Leftrightarrow-2x-51=0\)

\(\Leftrightarrow x=-\dfrac{51}{2}\)

c)\(\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-6\)

\(\Leftrightarrow\dfrac{2x-6x-3}{6}=\dfrac{x-36}{6}\)

\(\Leftrightarrow-4x-3=x-36\)

\(\Leftrightarrow-4x-3-x+36=0\)

\(\Leftrightarrow-5x+33=0\)

\(\Leftrightarrow x=\dfrac{33}{5}\)

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

\(\Leftrightarrow\dfrac{8+4x-6x}{12}=\dfrac{3-6x+3}{12}\)

\(\Leftrightarrow8-2x=6-6x\)

\(\Leftrightarrow8-2x-6+6x=0\)

\(\Leftrightarrow4x+2=0\)

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

Tính lại xem đúng không nha haha

16 tháng 1 2019

a) \(x-\dfrac{5x+2}{6}=\dfrac{7-3x}{4}\)

\(\Leftrightarrow\dfrac{24x}{24}-\dfrac{4\left(5x+2\right)}{24}=\dfrac{6\left(7-3x\right)}{24}\)

\(\Leftrightarrow24x-4\left(5x+2\right)=6\left(7-3x\right)\)

\(\Leftrightarrow24x-20x-8=42-18x\)

\(\Leftrightarrow4x-8=42-18x\)

\(\Leftrightarrow4x+18x=42+8\)

\(\Leftrightarrow22x=50\)

\(\Leftrightarrow x=\dfrac{25}{11}\)

Vậy S\(=\left\{\dfrac{25}{11}\right\}\)

a: \(\Leftrightarrow7\left(7-3x\right)+12\left(5x+2\right)=84\left(x+13\right)\)

\(\Leftrightarrow49-21x+60x+24=84x+1092\)

\(\Leftrightarrow39x-84x=1092-73\)

=>-45x=1019

hay x=-1019/45

b: \(\Leftrightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)

=>21x+63-14=20x+36-49x+63

=>21x+49=-29x+99

=>50x=50

hay x=1

c: \(\Leftrightarrow7\left(2x+1\right)-3\left(5x+2\right)=21x+63\)

=>14x+7-15x-6-21x-63=0

=>-22x-64=0

hay x=-32/11

d: \(\Leftrightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-17\cdot105\)

=>70x-105-30x-45=84x+63-1785

=>40x-150-84x+1722=0

=>-44x+1572=0

hay x=393/11

19 tháng 2 2022

a, msc 12.7=84 

Chuyển vế về =0 rồi làm

b,msc 28

c,làm tương tự

25 tháng 9 2021

1) ĐKXĐ: \(x^2+2x-3\ge0\Leftrightarrow\left(x+1\right)^2\ge4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1\ge2\\x+1\le-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\)

2) ĐKXĐ: \(2x^2+5x+3\ge0\Leftrightarrow2\left(x+\dfrac{5}{4}\right)^2\ge\dfrac{1}{8}\Leftrightarrow\left(x+\dfrac{5}{4}\right)^2\ge\dfrac{1}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{5}{4}\ge\dfrac{1}{4}\\x+\dfrac{5}{4}\le-\dfrac{1}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge-1\\x\le-\dfrac{3}{2}\end{matrix}\right.\)

3) ĐKXĐ: \(x-1>0\Leftrightarrow x>1\)

4) ĐKXĐ: \(x-3< 0\Leftrightarrow x< 3\)

5) ĐKXĐ: \(x+2< 0\Leftrightarrow x< -2\)

6) ĐKXĐ: \(2a-1>0\Leftrightarrow a>\dfrac{1}{2}\)

12 tháng 7 2017

\(a,\dfrac{x^2-2x}{x^2-4}=\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x+2}\)

b) \(\dfrac{x^2+5x+4}{x^2-1}=\dfrac{x^2+x+4x+4}{x^2-1}=\dfrac{\left(x+1\right)\left(x+4\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+4}{x-1}\)

c) \(\dfrac{x^4+4}{x\left(x^2+2\right)-2x^2-\left(x-1\right)^2-1}\)

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

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

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

\(=\dfrac{x^2+2+2x}{x-1}\)

Bài 2:

a) \(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)

\(=\dfrac{\left(2x+1\right)^2-\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}.\dfrac{5\left(2x-1\right)}{4x}\)

\(=\dfrac{8x}{\left(2x-1\right)\left(2x+1\right)}.\dfrac{5\left(2x-1\right)}{4x}\)

\(=\dfrac{10}{2x+1}\)

b) \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)

\(=\dfrac{1-2x+x^2}{x\left(x+1\right)}:\dfrac{1+x^2-2x}{x}\)

\(=\dfrac{1}{x+1}\)

c) Trong ngoặc giữa hai phân số là dấu gì vậy ?

14 tháng 7 2017

là dấu cộng