Bài 5:

a: \(A\left(1\right)=1^5-3\cdot1^4+1^2-5=1-3+1-5=-6\)

\(A\left(-1\right)=\left(-1\right)^5-3\cdot\left(-1\right)^4+\left(-1\right)^2-5\)

\(=-1-3\cdot1+1-5=-6-3+1=-8\)

b: \(B\left(1\right)=-1^4+2\cdot1^3-3\cdot1^2+4\cdot1+5\)

=-1+2-3+4+5

=1-3+4+5

=-2+4+5

=2+5

=7

\(B\left(-1\right)=-\left(-1\right)^4+2\cdot\left(-1\right)^3-3\cdot\left(-1\right)^2+4\cdot\left(-1\right)+5\)

\(=-1+2\cdot\left(-1\right)-3\cdot1-4+5\)

\(=-2-3=-5\)

c: \(C\left(1\right)=1+1^2+1^4+...+1^{100}\)

=1+1+...+1

=51

\(C\left(-1\right)=1+\left(-1\right)^2+\left(-1\right)^4+...+\left(-1\right)^{100}\)

=1+1+...+1

=51

d:

Từ x³ đến x¹⁰¹ thì có \(\dfrac{101-3}{2}+1=\dfrac{98}{2}+1=50\)(số hạng)

 \(D\left(1\right)=1+1^3+1^5+...+1^{101}\)

=1+1+...+1

=51

\(D\left(-1\right)=1+\left(-1\right)^3+\left(-1\right)^5+...+\left(-1\right)^{101}\)

\(=1-\left(1+1+...+1\right)\)

=1-50=-49

\(\dfrac{5}{42}:\dfrac{15}{7}+\dfrac{1}{2}\)

\(=\dfrac{5}{42}\times\dfrac{7}{15}+\dfrac{1}{2}\)

\(=\dfrac{1}{18}+\dfrac{1}{2}\)

\(=\dfrac{1}{18}+\dfrac{9}{18}=\dfrac{10}{18}=\dfrac{5}{9}\)

\(\rightarrow\) Chọn \(A.\dfrac{5}{9}\)

Câu 5:

\(\dfrac{5}{42}:\dfrac{15}{7}+\dfrac{1}{2}\)

\(=\dfrac{5}{42}\times\dfrac{7}{15}+\dfrac{1}{2}\)

\(=\dfrac{35}{630}+\dfrac{1}{2}=\dfrac{35}{630}+\dfrac{315}{630}=\dfrac{350}{630}=\dfrac{35}{63}=\dfrac{5}{9}\)

=>A

a: Trên tia Ax, ta có: AM<AB

nên M nằm giữa A và B

=>AM+MB=AB

=>MB+2,5=5

=>MB=2,5(cm)

b: Ta có: M nằm giữa A và B

mà MA=MB(=2,5cm)

nên M là trung điểm của AB

Bài 7: f(x)-h(x)=g(x)

=>h(x)=f(x)-g(x)

a: h(x)=f(x)-g(x)

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

\(=x^2+x+1-4+2x^3-x^4-7x^5\)

\(=-7x^5-x^4+2x^3+x^2+x-3\)

b: h(x)=f(x)-g(x)

\(=x^4+6x^3-4x^2+2x-1-x-3\)

\(=x^4+6x^3-4x^2+x-4\)

Bài 6:

f(1)=-3

=>\(a\cdot1+b=-3\)

=>a+b=-3(1)

f(2)=7

=>\(a\cdot2+b=7\left(2\right)\)

Từ (1),(2) ta có hệ phương trình:

\(\left\{{}\begin{matrix}a+b=-3\\2a+b=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b-2a-b=-3-7\\2a+b=7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-a=-10\\b=7-2a\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a=10\\b=7-2\cdot10=7-20=-13\end{matrix}\right.\)

\(6,2ha=62000m^2\)

\(4,02km^2=402ha\)

\(5200m^2< 5,2km^2\)

\(45,6dm^3< 46dm^3\)

\(139cm^3< 1,39dm^3\)

\(\dfrac{2}{5}dm^3< 0,6dm^3\)

Bài 4:

P(x)+Q(x)-R(x)

\(=6x^3-2x^2+3x-2-2x^3+3x^2-x+4-4x^3+2x-1\)

\(=x^2+4x+1\)

R(x)-P(x)-Q(x)

\(=-\left[P\left(x\right)+Q\left(x\right)-R\left(x\right)\right]\)

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

\(=-x^2-4x-1\)

Bài 3:

a: \(P\left(x\right)=-2x^4-7x+\dfrac{1}{2}-6x^4+2x^2-x\)

\(=\left(-2x^4-6x^4\right)+2x^2+\left(-7x-x\right)+\dfrac{1}{2}\)

\(=-8x^4+2x^2-8x+\dfrac{1}{2}\)

\(Q\left(x\right)=3x^3-x^4-5x^2+x^3-6x+\dfrac{3}{4}\)

\(=-x^4+\left(3x^3+x^3\right)+\left(-5x^2\right)-6x+\dfrac{3}{4}\)

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

b: P(x)+Q(x)

\(=-8x^4+2x^2-8x+\dfrac{1}{2}-x^4+4x^3-5x^2-6x+\dfrac{3}{4}\)

\(=-9x^4+4x^3-3x^2-14x+\dfrac{5}{4}\)

P(x)-Q(x)

\(=-8x^4+2x^2-8x+\dfrac{1}{2}+x^4-4x^3+5x^2+6x-\dfrac{3}{4}\)

\(=-7x^4-4x^3+7x^2-2x-\dfrac{1}{4}\)

Yêu cầu bạn để đúng lớp học!

$10\%+(-5):\frac{2}{10}$

$=0,1+(-5)\times\frac{10}{2}$

$=0,1+(-5)\times5$

$=0,1-25=-24,9$

4000m² = 0,4ha

1200dm² 20cm² = 12,002m²

45 phút = $\frac34$ giờ

32dm³ 800cm³ = 0,0328m³

\(4000m^2=0,4ha\)

45 phút=0,75 giờ

\(1200dm^220cm^2=12,002m^2\)

\(32dm^3800cm^3=0,0328m^3\)

Vì \(AN=\dfrac{2}{3}AC\)

nên \(S_{AMN}=\dfrac{2}{3}\times S_{AMC}\)

=>\(S_{AMC}=S_{AMN}:\dfrac{2}{3}=20:\dfrac{2}{3}=30\left(cm^2\right)\)

\(BM=\dfrac{1}{3}BC\)

=>\(CM=\dfrac{2}{3}BC\)

=>\(S_{AMC}=\dfrac{2}{3}\times S_{ABC}\)

=>\(S_{ABC}=30:\dfrac{2}{3}=45\left(cm^2\right)\)

Ở đây có mấy có gì bất thừơng ? 

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