a) P (x) = M (x) + N (x) = -2x² + 3 - 4x³ + 5x + 4x³ + 10 + 2x² - 3x

= ( -2x² + 2x² ) + ( 3 + 10 ) + ( -4x³ + 4x³ ) + ( 5x - 3x )

= 13 + 2x

Ta có P (x) = 13 + 2x

Đặt 13 + 2x = 0

=> 2x = - 13

=> x = \(\dfrac{-13}{2}\)

Vậy nghiệm của đa thức P (x) là \(\dfrac{-13}{2}\)

b) Q (x) = M (x) - N (x) 

=> Q (x) =  -2x² + 3 - 4x³ + 5x - ( 4x³ + 10 + 2x² - 3x)

=  -2x² + 3 - 4x³ + 5x - 4x³ - 10 - 2x² + 3x

= ( -2x² - 2x² ) + ( 3 - 10 ) + ( -4x³ - 4x³ ) + ( 5x + 3x )

= -4x² - 7 - 8x³ + 8x

a) P (x) = M (x) + N (x) = -2x² + 3 - 4x³ + 5x + 4x³ + 10 + 2x² - 3x

= ( -2x² + 2x² ) + ( 3 + 10 ) + ( -4x³ + 4x³ ) + ( 5x - 3x )

= 13 + 2x

Ta có P (x) = 13 + 2x

Đặt 13 + 2x = 0

=> 2x = - 13

=> x = −132−132

b) Q (x) = M (x) - N (x) 

=> Q (x) =  -2x² + 3 - 4x³ + 5x - ( 4x³ + 10 + 2x² - 3x)

=  -2x² + 3 - 4x³ + 5x - 4x³ - 10 - 2x² + 3x

= ( -2x² - 2x² ) + ( 3 - 10 ) + ( -4x³ - 4x³ ) + ( 5x + 3x )

= -4x² - 7 - 8x³ + 8x

Chọn A

a: Xét ΔABH vuông tại H và ΔMBH vuông tại H có 

HB chung

HA=HM

Do đó: ΔABH=ΔMBH

S BCD=S DHB+S DHC

=1/2*HD*HB+1/2*HD*HC

=1/2*HD*12=6*HD

=6*1/3*AH=2AH

S ABH=1/2*BH*AH=1/2*4*AH=2AH

=>S ABH=S DBC

1: CN=3/5AC

=>S BNC/S BAC=3/5

=>S BNC=490*3/5=294cm²

2: EC=1/4BC

=>S AEC=1/4*S ABC

=>S ABC=4*117=468m²

Bài 4: 

45 phút=3/4 giờ

Bài 5: 

x+3/10=-2/5

=>x=-2/5-3/10=-7/10

Bài 6: 

\(=\dfrac{3^9\cdot5^{11}\left(5-3^2\right)}{5^{10}\cdot3^{10}}=\dfrac{5\cdot\left(-4\right)}{3}=-\dfrac{20}{3}\)

= 20/3

= 20/3

\(m_{NaCl}=\dfrac{150.10}{100}=15\left(g\right)\\ m_{H_2O}=150-15=135\left(g\right)\)

Đong 15 gam NaCl khan, 135 gam nước cho sẵn vào cốc nước. Sau đó hóa tan 15 gam NaCl vào nước, dùng đũa thủy tinh khuấy đều thu được dd như theo yêu cầu của đề bài

Lời giải:

Vì $(d_2)$ có hệ số góc là $3$ nên $m=3$

$(d_2)$ cắt trục hoành tại điểm có hoành độ $-2$, nghĩa là $(d_2)$ đi qua $(-2;0)$

$\Rightarrow 0=m.(-2)+n=3(-2)+n$

$\Rightarrow n=6$

Vậy $m=3; n=6$

a) Ta có: (x-5)(x+5)=2x-5

\(\Leftrightarrow x^2-25-2x+5=0\)

\(\Leftrightarrow x^2-2x-20=0\)

\(\Leftrightarrow x^2-2x+1=21\)

\(\Leftrightarrow\left(x-1\right)^2=21\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=\sqrt{21}\\x-1=-\sqrt{21}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{21}+1\\x=-\sqrt{21}+1\end{matrix}\right.\)

Vậy: \(S=\left\{\sqrt{21}+1;-\sqrt{21}+1\right\}\)

b) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

Ta có: \(\dfrac{x}{x-1}-\dfrac{2x}{x^2-1}=0\)

\(\Leftrightarrow\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}=0\)

Suy ra: \(x^2+x-2x=0\)

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

\(\Leftrightarrow x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=1\left(loại\right)\end{matrix}\right.\)

Vậy: S={0}

c)ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

Ta có: \(\dfrac{2x}{x+1}-\dfrac{4}{x^2-1}=\dfrac{2x-5}{x-1}\)

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

Suy ra: \(2x^2-2x-4=2x^2+2x-5x-5\)

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

\(\Leftrightarrow x+1=0\)

hay x=-1(loại)

Vậy: \(S=\varnothing\)

d)ĐKXĐ: \(x\notin\left\{-2;-3\right\}\)

Ta có: \(\dfrac{5\left(x-2\right)}{x+2}-\dfrac{2\left(x-3\right)}{x+3}=3\)

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

Suy ra: \(5\left(x^2+x-6\right)-2\left(x^2-x-6\right)=3\left(x^2+5x+6\right)\)

\(\Leftrightarrow5x^2+5x-30-2x^2+2x+12=3x^2+15x+18\)

\(\Leftrightarrow3x^2+7x-18-3x^2-15x-18=0\)

\(\Leftrightarrow-8x-36=0\)

\(\Leftrightarrow-8x=36\)

hay \(x=-\dfrac{9}{2}\)(thỏa ĐK)

Vậy: \(S=\left\{-\dfrac{9}{2}\right\}\)

e) Ta có: \(\left|2x-10\right|=-3x\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-10=-3x\left(x\ge5\right)\\-2x+10=-3x\left(x< 5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3x-10=0\\-2x+3x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=10\\x=-10\left(nhận\right)\end{matrix}\right.\)

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

Vậy: S={-10}