\(0,3:2,5=3:25\)

\(4\dfrac{2}{5}:1\dfrac{1}{3}=\dfrac{22}{5}:\dfrac{4}{3}=33:10\)

\(-3,2:1\dfrac{2}{7}=\dfrac{-16}{5}:\dfrac{9}{7}=112:45\)

a

b) Xét phương trình hoành độ giao điểm của (P) và (d'):

\(x^2=x-2m+1\)

\(\Leftrightarrow x^2-x+2m-1=0\)

\(\Delta=\left(-1\right)^2-4.1.\left(2m-1\right)=5-8m\)

Để (d') cắt (P) tại 2 điểm phân biệt: \(\Delta>0\Leftrightarrow5>8m\Leftrightarrow m< \dfrac{5}{8}\)

Theo định lí Vi-et:

\(\left\{{}\begin{matrix}x_1+x_2=1\\x_1.x_2=2m-1\end{matrix}\right.\)

Theo bài: \(x_1^2+x_2^2=7\)

\(\Leftrightarrow x_1^2+2x_1x_2+x_2^2-2x_1x_2=7\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=7\)

\(\Leftrightarrow1^2-2\left(2m-1\right)=7\)

\(\Leftrightarrow1-4m+2=7\)

\(\Leftrightarrow-4m=4\Leftrightarrow m=-1\left(tmm< \dfrac{5}{8}\right)\)

Vậy m = -1 là giá trị cần tìm

a) 

b) Đường thẳng (d') cắt (P) ta có phương trình hoành độ giao điểm là:

\(x^2=x-2m+1\)

\(\Leftrightarrow x^2-x+2m-1=0\)

\(\Delta=\left(-1\right)^2-4\cdot1\cdot\left(2m-1\right)=1-8m+4=-8m+5\)

\(\Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{1+\sqrt{5-8m}}{2}\\x_2=\dfrac{1-\sqrt{5-8m}}{2}\end{matrix}\right.\left(đk:m\le\dfrac{5}{8}\right)\)

Mà: \(x^2_1+x^2_2=7\)

\(\Leftrightarrow\left(\dfrac{1+\sqrt{5-8m}}{2}\right)^2+\left(\dfrac{1-\sqrt{5-8m}}{2}\right)^2=7\)

\(\Leftrightarrow\dfrac{1+2\sqrt{5-8m}+5-8m}{4}+\dfrac{1-2\sqrt{5-8m}+5-8m}{4}=7\)

\(\Leftrightarrow\dfrac{6+2\sqrt{5-8m}-8m+6-2\sqrt{5-8m}-8m}{4}=7\)

\(\Leftrightarrow12-16m=28\)

\(\Leftrightarrow-16m=16\)

\(\Leftrightarrow m=-1\left(tm\right)\)

Vậy: .... 

\(A=-x\left(x-6\right)+7\)

\(=-x^2+6x+7\)

\(=-\left(x^2-6x-7\right)\)

\(=-\left(x^2-6x+9-16\right)\)

\(=-\left(x-3\right)^2+16\le16\forall x\)

Dấu '=' xảy ra khi x=3

-x-2/3=-6/7

-x      =-6/7+2/3

-x      =-18/21+14/21

-x      =-4/21

=> x=4/21

=> -x= -(2/3 + 6/7)
=> -x= -32/21

=> x=32/21

4/9-5/18 = 8/18 - 5/18 = 3/18 = 1/6

3/5x5/7 = 3/7

4/7:2/7 = 4/7 x 7/2 = 2

4/9-5/18= 8/18 - 5/18 = 3/18 = 1/6

 3/5x5/7=15/35 = 3/7    

4/7:2/7=4/7 x 7/2 = 4/2 = 2

a: =>2x^2-2x+2x-2-2x^2-x-4x-2=0

=>-5x-4=0

=>x=-4/5

b: =>6x^2-9x+2x-3-6x^2-12x=16

=>-19x=19

=>x=-1

c: =>48x^2-12x-20x+5+3x-48x^2-7+112x=81

=>83x=83

=>x=1

a: =>(x-5)(x+5)+(x-5)(3x-15)=0

=>(x-5)(x+5+3x-15)=0

=>(x-5)(4x-10)=0

=>x=5 hoặc x=5/2

c: =>x^3-3x^2+2x^2-6x-8x+24=0

=>(x-3)(x^2+2x-8)=0

=>(x-3)(x+4)(x-2)=0

=>\(x\in\left\{3;-4;2\right\}\)

`A(x)=-2x^2+5x+7=0`

`=> -(2x^2-5x+7)=0`

`=> -(2x^2-2x-7x+7)=0`

`=> -[(2x^2-2x)-(7x-7)]=0`

`=> -[2x(x-1)-7(x-1)]=0`

`=> -[(2x-7)(x-1)]=0`

`=> -(2x-7)(x-1)=0`

`=> (2x-7)(x+1)=0`

`=>`\(\left[{}\begin{matrix}2x-7=0\\x+1=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=7\\x=-1\end{matrix}\right.\)

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

Vậy, nghiệm của đa thức là `x={7/2; -1}.`

a)\(3x-\dfrac{2}{5}=0=>3x=\dfrac{2}{5}=>x=\dfrac{2}{15}\)

b)\(\left(x-3\right)\left(2x+8\right)=0=>\left[{}\begin{matrix}x-3=0\\2x=-8\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)

c)\(3x^2-x-4=0=>3x^2+3x-4x-4=0=>\left(3x-4\right)\left(x+1\right)=0\)

\(=>\left[{}\begin{matrix}3x=4\\x+1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-1\end{matrix}\right.\)