Bài 2: 

d) Ta có: \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)

\(=\sqrt{5}+1+\sqrt{5}-1\)

\(=2\sqrt{5}\)

e) Ta có: \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)

\(=3+\sqrt{2}-3+\sqrt{2}\)

\(=2\sqrt{2}\)

c) thay x=1 vào đa thức f(x) ta có:  f(1)=4.1^3-1^2+2.1-5

                                                             =4-2+2-5

                                                             =- 1

    vậy 1 k phải là nghiệm của đa thức f(x)

MÌNH CHỈ LÀM ĐƯỢC C THÔI HOK TỐT

làm sai nha chỗ nào là 1 thì thay bằng -1 nha kq sẽ ra nha

a) \(f\left(x\right)+g\left(x\right)-h\left(x\right)\)

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

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

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

\(=10x^3-0+3x-14\)

\(=10x^3+3x-14\)

phắc

\(2,\\ a,x=36\Leftrightarrow P=\dfrac{6+1}{6-2}=\dfrac{7}{4}\\ b,x=6-2\sqrt{5}\Leftrightarrow\sqrt{x}=\sqrt{5}-1\\ \Leftrightarrow P=\dfrac{\sqrt{5}-1+1}{\sqrt{5}-1-2}=\dfrac{\sqrt{5}}{\sqrt{5}-3}=\dfrac{5-3\sqrt{5}}{2}\\ c,x=\dfrac{2}{2+\sqrt{3}}=4-2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}-1\\ \Leftrightarrow P=\dfrac{\sqrt{3}-1+1}{\sqrt{3}-1-2}=\dfrac{\sqrt{3}}{\sqrt{3}-3}=\dfrac{3\left(\sqrt{3}+1\right)}{-6}=\dfrac{-\sqrt{3}-1}{2}\)

h: Ta có: \(x-7\sqrt{x}+10=0\)

nên x=25

Thay x=25 vào P, ta được:

\(P=\dfrac{5+1}{5-2}=\dfrac{6}{3}=2\)

d) \(x^2=a\left(a\ge0\right)\)

\(\Rightarrow x=\sqrt{a}\)

e) \(x^2=\dfrac{4}{9}\)

\(\Rightarrow x^2=\left(\pm\dfrac{2}{3}\right)^2\)

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

f) \(x^2-\dfrac{16}{25}=0\)

\(\Rightarrow x^2=\dfrac{16}{25}\)

\(\Rightarrow x^2=\left(\pm\dfrac{4}{5}\right)^2\)

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

g) \(x^2-\dfrac{7}{36}=0\)

\(\Rightarrow x^2=\dfrac{7}{36}\)

\(\Rightarrow x^2=\left(\pm\sqrt{\dfrac{7}{36}}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{7}{36}}\\x=-\sqrt{\dfrac{7}{36}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{7}}{6}\\x=-\dfrac{\sqrt{7}}{6}\end{matrix}\right.\)

h) Ta có: \(x^2\ge0\forall x\)

\(\Rightarrow x^2+1\ge1>0\forall x\)

mà \(x^2+1=0\)

nên không tìm được giá trị nào của x thoả mãn đề bài.

\(a,Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\\ b,Fe_2O_3+SO_3\rightarrow Không.p.ứ\\ c,Na_2O+H_2O\rightarrow2NaOH\\ d,Fe_3O_4+8HCl\rightarrow2FeCl_3+FeCl_2+4H_2O\\ e,SO_3+H_2O\rightarrow H_2SO_4\\ f,2NaOH\left(dư\right)+CO_2\rightarrow Na_2CO_3+H_2O\\ g,P_2O_5+3H_2O\rightarrow2H_3PO_4\\ h,CaO+CO_2\rightarrow CaCO_3\)

a: Ta có: \(\sqrt{75}-\sqrt{5\dfrac{1}{3}}+\dfrac{9}{2}\sqrt{2\dfrac{2}{3}}+2\sqrt{27}\)

\(=5\sqrt{3}+\dfrac{4}{3}\sqrt{3}+3\sqrt{6}+6\sqrt{3}\)

\(=\dfrac{37}{3}\sqrt{3}+3\sqrt{6}\)

c: Ta có: \(\left(\sqrt{12}+2\sqrt{27}\right)\cdot\dfrac{\sqrt{3}}{2}-\sqrt{150}\)

\(=\left(2\sqrt{3}+6\sqrt{3}\right)\cdot\dfrac{\sqrt{3}}{2}-5\sqrt{6}\)

\(=12-5\sqrt{6}\)

\(e,=\dfrac{\left(3+\sqrt{2}\right)\left(2\sqrt{2}+1\right)}{7}-\sqrt{\dfrac{\left(\sqrt{2}+1\right)^2}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}}\\ =\dfrac{7\sqrt{2}+7}{7}-\dfrac{\sqrt{2}+1}{1}=\sqrt{2}+1-\sqrt{2}-1=0\)

\(f,=\sqrt{\dfrac{\left(2\sqrt{3}-3\right)^2}{\left(2\sqrt{3}-3\right)\left(2\sqrt{3}+3\right)}}\left(2+\sqrt{3}\right)\\ =\dfrac{\left(2\sqrt{3}-3\right)\left(2+\sqrt{3}\right)}{\sqrt{3}}\\ =\dfrac{\sqrt{3}}{\sqrt{3}}=1\)

\(h,=\sqrt{\dfrac{\left(3\sqrt{5}-1\right)\left(2\sqrt{5}-3\right)}{20-9}}\left(\sqrt{2}+\sqrt{10}\right)\\ =\sqrt{\dfrac{2\left(33-11\sqrt{5}\right)}{11}}\left(\sqrt{5}+1\right)\\ =\sqrt{\dfrac{22\left(3-\sqrt{5}\right)}{11}}\left(\sqrt{5}+1\right)\\ =\sqrt{6-2\sqrt{5}}\left(\sqrt{5}+1\right)=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=4\)