a) M(x) = A(x) - 2B(x) + C(x)

\(\Leftrightarrow\)M(x) = 2x⁵ - 4x³ + x² - 2x + 2 - 2(x⁵ - 2x⁴ + x² - 5x + 3) + x⁴ + 4x³ + 3x² - 8x + \(4\frac{3}{16}\)

\(\Leftrightarrow\)M(x) = 2x⁵ - 4x³ + x² - 2x + 2 - 2x⁵ - 4x⁴ - 2x² + 10x - 6 + x⁴ + 4x³ + 3x² - 8x + \(4\frac{3}{16}\)

\(\Leftrightarrow\)M(x) = (2x⁵ - 2x⁵) + (-4x³ + 4x³) + (x² - 2x² + 3x²) + (-2x + 10x - 8x) + (2 - 6 + \(4\frac{3}{16}\))

\(\Leftrightarrow\)M(x) = 2x² + \(\frac{3}{16}\)

b) Thay \(x=-\sqrt{0,25}\)vào M(x), ta được:

\(M\left(x\right)=2\left(-\sqrt{0,25}\right)^2+\frac{3}{16}\)

\(M\left(x\right)=2.0,25+\frac{3}{16}\)

\(M\left(x\right)=0,5+\frac{3}{16}\)

\(M\left(x\right)=\frac{11}{16}\)

c) Ta có : \(x^2\ge0\)

\(\Leftrightarrow2x^2+\frac{3}{16}\ge\frac{3}{16}\)

Vậy để \(M\left(x\right)=0\Leftrightarrow x\in\varnothing\)

a) Ta có:

\(M\left(x\right)=A\left(x\right)-2.B\left(x\right)+C\left(x\right)\)

\(=\left(2x^5-4x^3+x^2-2x+2\right)-2.\left(x^5-2x^4+x^2-5x+3\right)+\left(x^4+3x^3+3x^2-8x+4\frac{3}{16}\right)\)

\(=2x^5-4x^3+x^2-2x+2-2x^5+4x^4-2x^2+10x-6+x^4+4x^3+3x^2-8x+\frac{67}{16}\)

\(=\left(2x^5-2x^5\right)+\left(4x^4+x^4\right)+\left(-4x^3+4x^3\right)+\left(x^2-2x^2+3x^2\right)+\left(-2x+10x-8x\right)+\left(2-6+\frac{67}{16}\right)\)

\(=0+5x^4+0+2x^2+0+\frac{3}{16}\)

\(=5x^4+2x^2+\frac{3}{16}\)

b) Thay  \(x=-\sqrt{0,25}=-0,5\); ta có:

\(M\left(-0,5\right)=5.\left(-0,5\right)^4+2.\left(-0,5\right)^2+\frac{3}{16}\)

\(=5.0,0625+2.0,25+\frac{3}{16}\)

\(=\frac{5}{16}+\frac{8}{16}+\frac{3}{16}=\frac{16}{16}=1\)

c) Ta có:

\(x^4\ge0\) với mọi x

\(x^2\ge0\) với mọi x

\(\Rightarrow5x^4+2x^2+\frac{3}{16}>0\) với mọi x

Do đó không có x để M(x)=0

Lời giải:

1.

\(M(x)=A(x)-2B(x)+C(x)\)

\(2x^5 – 4x^3 + x^2 – 2x + 2-2(x^5 – 2x^4 + x^2 – 5x + 3)+ (x^4 + 4x^3 + 3x^2 – 8x + \frac{43}{16})\)

\(=5x^4+2x^2-\frac{21}{16}\)

2.

Khi $x=-\sqrt{0,25}=-0,5$ thì:

\(M(x)=5.(-0,5)^4+2(-0,5)^2-\frac{21}{16}=\frac{-1}{2}\)

3)

$M(x)=0$

$\Leftrightarrow 5x^4+2x^2-\frac{21}{16}=0$

$\Leftrightarrow 80x^4+32x^2-21=0$

$\Leftrightarrow 4x^2(20x^2-7)+3(20x^2-7)=0$

$\Leftrightarrow (4x^2+3)(20x^2-7)=0$

Vì $4x^2+3>0$ với mọi $x$ thực nên $20x^2-7=0$

$\Rightarrow x=\pm \sqrt{\frac{7}{20}}$

Đây chính là giá trị của $x$ để $M(x)=0$

a,

C(x)=-3x^4-2x^3+x^2+x+5

lop 7 lam gi co nghiem voi da thuc ha ban

Đề thi HSG lớp 7 đó bạn

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

\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)

\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)

\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)

\(\Leftrightarrow-7x+12x=20+2\)

\(\Leftrightarrow5x=22\)

\(\Rightarrow x=\dfrac{22}{5}\)

tick cho mk nha

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

\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)

\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)

\(\Leftrightarrow10x^2-19x-33=0\)

\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)

\(x_1=3;x_2=\dfrac{-11}{10}\)

Tick cho mk nha

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

\(\Leftrightarrow21x-15x^2-35+25x-4x+15x^2-4=4\)

\(\Leftrightarrow42x-39=4\)

\(\Leftrightarrow42x=4+39\)

\(\Leftrightarrow42x=43\)

\(\Rightarrow x=\dfrac{43}{42}\)

Tick cho mk nha

Giups mik vs

a: =2x^5-15x^3-x^2-2x^5-x^3=-16x^3-x^2

b: =x^3+3x^2-2x-3x^2-9x+6

=x^3-11x+6

c: \(=\dfrac{4x^3+2x^2-6x^2-3x-2x-1+5}{2x+1}\)

\(=2x^2-3x-1+\dfrac{5}{2x+1}\)

a) \(6x^3\left(\dfrac{1}{3}x^2-\dfrac{5}{2}-\dfrac{1}{6}\right)-2x^5-x^3\)

\(=6x^3\left(\dfrac{1}{3}x^2-\dfrac{16}{6}\right)-2x^5-x^3\)

\(=2x^5-16x^3-2x^5-x^3\)

\(=-17x^3\)

b) \(\left(x+3\right)\left(x^2+3x-2\right)\)

\(=x^3+3x^2-2x+3x^2+9x-6\)

\(=x^3+6x^2+7x-6\)

c) \(\left(4x^3-4x^2-5x+4\right):\left(2x+1\right)\)

\(=2x^2+4x^3-2x-4x^2-\dfrac{5}{2}-5x+\dfrac{2}{x}+4\)

\(=4x^3-2x^2-7x+\dfrac{2}{x}+\dfrac{3}{2}\)