\(x=\left(\dfrac{1}{2}\right)^3:\left(\dfrac{1}{2}\right)=\left(\dfrac{1}{2}\right)^{3-1}=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)

1/4

\(\dfrac{1}{4}\)

Ta có: \(a+b=5\Rightarrow a=5-b\)

Thay \(a=5-b\) vào \(2a-b=4\) ta có:

\(2\cdot\left(5-b\right)-b\)

\(\Rightarrow10-2b-b=4\)

\(\Rightarrow10-3b=4\)

\(\Rightarrow3b=10-4\)

\(\Rightarrow3b=6\)

\(\Rightarrow b=\dfrac{6}{3}=2\)

Lúc này ta tìm được \(a\):

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

Vậy: \(a=3,b=2\)

10.

\(H\left(x\right)=-5x^4+10x^3-15x+1\)

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

\(=-5x.0+1\)

\(=1\)

9.

\(P\left(x\right)-Q\left(x\right)=\left(1-a\right)x^3+x^2+x-6\)

\(P\left(x\right)-Q\left(x\right)\) là đa thức bậc 3 khi và chỉ khi \(1-a\ne0\)

\(\Rightarrow a\ne1\)

Bài 2: 

b: Ta có: \(B=\dfrac{15-5\sqrt{x}}{x-5\sqrt{x}+6}+\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)

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

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-2}=1\)

giúp em với mng ơi

\(b,\widehat{K_1}=\widehat{KNQ}+\widehat{NQK}=41+90=131\left(góc.ngoài\right)\)

\(d,NH\perp PQ;PQ//NK\Rightarrow NH\perp NK\Rightarrow\widehat{HNK}=90\)

\(\widehat{NKQ}=180-\widehat{K_1}=180-131=49\)

\(\widehat{HAQ}=180-\widehat{HNK}-\widehat{NKQ}=180-90-49=41\)

\(lim\left(\sqrt[3]{n^3+4}-\sqrt[3]{n^3-1}\right)\)

\(=lim\left(\sqrt[3]{1+\dfrac{4}{n^3}}-\sqrt[3]{1-\dfrac{1}{n^3}}\right)=\sqrt[3]{1}-\sqrt[3]{1}=0\)