a, \(E=4x^2+6x+5=4\left(x^2+\frac{2.3}{4}x+\frac{9}{16}-\frac{9}{16}\right)+5\)

\(=4\left(x+\frac{3}{4}\right)^2+\frac{11}{4}\ge\frac{11}{4}>0\forall x\)

Vậy ta có đpcm 

b, \(F=2x^2-3x+7=2\left(x^2-\frac{2.3}{4}x+\frac{9}{16}-\frac{9}{16}\right)+7\)

\(=2\left(x-\frac{3}{4}\right)^2+\frac{47}{8}\ge\frac{47}{8}>0\forall x\)

Vậy ta có đpcm 

c, \(K=5x^2-4x+1=5\left(x^2-\frac{2.2}{5}x+\frac{4}{25}-\frac{4}{25}\right)+1\)

\(=5\left(x-\frac{2}{5}\right)^2+\frac{1}{5}\ge\frac{1}{5}>0\forall x\)

Vậy ta có đpcm 

d, \(Q=3x^2+2x+5=3\left(x^2+\frac{2}{3}x+\frac{1}{9}-\frac{1}{9}\right)+5\)

\(=3\left(x+\frac{1}{3}\right)^2+\frac{14}{3}\ge\frac{14}{3}>0\forall x\)

Vậy ta có đpcm 

6x(5x+3)+3x(1+10x)=7

=> 30x²+18+3x+30x²=7

=> 60x²+3x+18=7

=>60x²+3x=-11

=> ......

a/ 3x(12x-4)-9x(4x-3)

=36x²-12x-36x²+27x

=(36x²-36x²)-12x+27x

=15x

b/ x(5-2x)+2x(x-1)

=5x-2x²+2x²-2x

=(5x-2x)-(-2x²+2x²)

=3x

c/ 5x(12x+7)-3x(20x-5)

=60x²+35x-60x²+15x

=(60x²-60x²)+(35x+15x)\

=50x

d/ 3x(2x-7)+2x(5-3x)

=6x²-21x+10x-6x²

=(6x²-6x²)+(10x-21x)

=-11x

e/ đề sai hay sao ý ra số to lắm @@

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

b)\(\left(3x-5\right)\left(5x-7\right)+\left(5x+1\right)\left(2-3x\right)=4\)

\(\Leftrightarrow15x^2-46x+35-15x^2+7x+2-4=0\)

\(\Leftrightarrow33-39x=0\Leftrightarrow33=39x\Leftrightarrow x=\frac{33}{39}\)

a) \(x\left(x-3\right)-2x+6=0\)

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

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

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

b) \((3x-5)(5x-7)+(5x+1)(2-3x)=4\)

\(15x^2-46x+35+10x-15x^2+2-3x-4=0\)

\(33-39x=0\)

\(3\left(11-13x\right)=0\)

\(11-13x=0\)

\(13x=11\)

\(x=\frac{11}{13}\)

lớp11

(8x - 3)(3x + 2) - (4x + 7)(x + 4) = (2x + 1)(5x - 1) - 33

=> 24x² + 16x - 9x - 6 - 4x² - 16x - 7x - 28 = 10x² - 2x + 5x - 1 - 33

=> 10x² - 19x = 0

=> x(10x - 19) = 0

=> x = 0

hoặc 10x - 19 = 0 => x = 19/10

                                                               Vậy x = 0, x = 19/10

+ \(72:\left(x-15\right)=8\)

\(x-15=9\)

\(x=24\)

+ \(\dfrac{3}{90}-3x=3\)

\(3\left(\dfrac{1}{90}-x\right)=3\)

\(\dfrac{1}{90}-x=1\)

\(x=\dfrac{1}{90}-1=\dfrac{-89}{90}\)

`@` `\text {Ans}`

`\downarrow`

`(8x-3)(3x+2)-(4x+7)(x+4)=(2x+1)(5x-1)-33`

`\Leftrightarrow 8x(3x+2) -3(3x+2) - 4x(x+4) + 7(x+4) = 2x(5x-1) + 5x-1 - 33`

`\Leftrightarrow 24x^2 + 16x - 9x - 6 - 4x^2 - 16x - 7x - 28 = 10x^2 - 2x + 5x - 1 - 33`

`\Leftrightarrow 20x^2 -16x - 34 = 10x^2 + 3x - 34`

`\Leftrightarrow 20x^2 - 16x - 34 - 10x^2 - 3x + 34 = 0`

`\Leftrightarrow 10x^2 - 19x = 0`

`\Leftrightarrow x(10x - 19)=0`

`\Leftrightarrow `\(\left[{}\begin{matrix}x=0\\10x-19=0\end{matrix}\right.\)

`\Leftrightarrow `\(\left[{}\begin{matrix}x=0\\10x=19\end{matrix}\right.\)

`\Leftrightarrow `\(\left[{}\begin{matrix}x=0\\x=\dfrac{19}{10}\end{matrix}\right.\)

Vậy, `x={0; 19/10}.`

Với bài này bn áp dụng bài phần tử của tập hợp nhé! 

(8-4):6=129

Gọi 129 là x

X-7=59

Gọi  59 làc

Vậy phần bài này là phần tử

Đs 78/9

\(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)-33\)

\(\Leftrightarrow24x^2+16x-9x-6-4x^2-16x-7x-28=10x^2-2x+5x-1-33\)

\(\Leftrightarrow10x^2-19x=0\)

\(\Leftrightarrow x\left(10x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\10x-19=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{10}\end{matrix}\right.\)

minh moi hoc lop 5 thoi

1) 3x - 6=  5x + 2

5x - 3x = -6 - 2

2x = -8

x = -4

2) 15 - x = 4x - 5

4x + x = 15 + 5

5x = 20

x = 4

Tương tự như trên 

1) 3x - 6 = 5x + 2

  3x - 5x = 2 + 6

        -2x = 8

           x = 8 : (-2)

           x = -4

2) 15 - x = 4x - 5

    -x - 4x = -5 - 15

         -5x = -20

            x = -20 : (-5)

            x = 4

Ta có : A = \(\frac{-5x}{x^2-3x+\frac{9}{4}+\frac{31}{4}}\)= \(\frac{-5x}{\left(x-\frac{3}{2}\right)^2+\frac{31}{4}}\)Vì \(\left(x-\frac{3}{2}\right)^2\)>0 hoặc =0 , khi công thêm \(\frac{31}{4}\)thì Mẫu số luôn lớ hơn hoặc bằng 0. Mà -5x luôn bé hơn hoặc bằng 0

Vì vậy biểu thức A luôn âm 

cảm ơn bạn rất nhều !!!