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    2x^3y^2 - 5x^3y^3 + 6x^3y^2 - 8x^3y^2

=  ( \(2x^3y^2+6x^3y^2-8x^3y^2\)  ) - \(5x^3y^3\)

=  0- \(5x^3y^3\)

= -5x^3y^3

Thay x=-1 và y=1 vào biểu thức rút gọn trên:

-5 . (-1)^3 . 1^3=  -5 

Vậy giá trị biểu thức = -5

  2x^3y^2 - 5x^3y^3 + 6x^3y^2 - 8x^3y^2

\(\left(2x^3y^2+6x^3y^2-8x^3y^2\right)\)-\(5x^3y^3\)

=  0-\(5x^3y^3\)

= -5x^3y^3

Thay x=-1 và y=1 vào biểu thức rút gọn trên:

-5 . (-1)^3 . 1^3=  5 

Vậy giá trị biểu thức = 5

Áp dụng định lí pytago trong \(\Delta MNP\) vuông tại \(M\) có:

\(\Rightarrow MP=\sqrt{NP^2-MN^2}=\sqrt{7,5^2-4,5^2}=6cm\)

\(\Rightarrow D\)

Chọn D nhé bạn

1) \(\left(x-2\right)\left(4x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\4x=6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{3}{2}\end{matrix}\right.\)

2) \(\left(2x^2+5\right)\left(5-10x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^2+5=0\\5-10x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^2=-5\\10x=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=-\dfrac{5}{2}\left(\text{vô lí}\right)\\x=\dfrac{1}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow\left(x-3\right)\left(2x+6\right)-\left(4+3x\right)\left(3-x\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+6\right)+\left(4+3x\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left[\left(2x+6\right)+\left(4+3x\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(5x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\5x+10=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\5x=-10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

4) \(\left(4x-3\right)\left(x-5\right)=x^2-16\)

\(\Leftrightarrow\left(4x^2-20x-3x+15\right)-\left(x^2-16\right)=0\)

\(\Leftrightarrow4x^2-23x+15-x^2+16=0\)

\(\Leftrightarrow3x^2-23x+31=0\)

\(\Delta=\left(-23\right)^2-4\cdot3\cdot31=157>0\)

\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{-23+\sqrt{157}}{6}\\x_2=\dfrac{-23-\sqrt{157}}{6}\end{matrix}\right.\)

5) \(\left(3x+1\right)^2=x^2-8x+16\)

\(\Leftrightarrow\left(3x+1\right)^2=\left(x-4\right)^2\)

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

\(\Leftrightarrow\left[\left(3x+1\right)-\left(x-4\right)\right]\left[\left(3x+1\right)+\left(x-4\right)\right]=0\)

\(\Leftrightarrow\left(2x+5\right)\left(4x-3\right)=0\)

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

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

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

1: =>x-2=0 hoặc 4x-6=0

=>x=2 hoặc x=3/2

2: =>5-10x=0

=>10x=5

=>x=1/2

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

=>(x-3)(2x+6+3x+4)=0

=>(x-3)(5x+10)=0

=>x=3 hoặc x=-2

4: =>4x^2-20x-3x+15-x^2+16=0

=>3x^2-23x+31=0

=>\(x=\dfrac{23\pm\sqrt{157}}{6}\)

5: =>(3x+1)^2-(x-4)^2=0

=>(3x+1+x-4)(3x+1-x+4)=0

=>(4x-3)(2x+5)=0

=>x=3/4 hoặc x=-5/2

ĐỀ ĐÂY A

8700

8700

8700

a, -3,5

b, -0,35

c, -3,07

d, -0,63

a, -3,5

b, -0,35

c, -3,07

d, -0,63

a, -3,5

b, -0,35

c, -0,37

d,-0,63

1+2+3+4+5+6+7+8+9+10=55       Ko chắc:=))

bằng 55

BN Nguyễn Ngọc Diệp sai r = 55 mà