a, c.Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath

cau a,b,c thay no co chung 1 dang do la

\(\sqrt[3]{a+m}+\sqrt[3]{a-m}\)

dang nay co 2 cach

C1: nhanh kho nhin de sai

VD: cau B

\(B^3=40+3\sqrt[3]{\left(20+14\sqrt{2}\right)\left(20-14\sqrt{2}\right)}\left(B\right)\)

B^3=40+3(2)(B)

B^3=40+6B

B=4

C2: hoi dai nhung de nhin

dat \(a=\sqrt[3]{20+14\sqrt{2}};b=\sqrt[3]{20-14\sqrt{2}}\)

de thay B=a+b

            ab=2

            a^3+b^3=40

suy ra B^3=a^3+b^3+3ab(a+b)

B^3=40+6B

B=4

giai tuong tu

con co cach nay nhung it su dung vi kho tim

C3: dua ve tong lap phuong

VD:cau B

 \(20+14\sqrt{2}=\left(2+\sqrt{2}\right)^3\)

\(20-14\sqrt{2}=\left(2-\sqrt{2}\right)^3\)

de thay

B=4

cau d)

dung CT nay

\(\sqrt[m]{a}=\sqrt[m\cdot n]{\left(a\right)^n}\)

ap dung vao bai

\(\sqrt[3]{2\sqrt{3}-4\sqrt{2}}=\sqrt[6]{\left(2\sqrt{3}-4\sqrt{2}\right)^2}=\sqrt[6]{44-16\sqrt{6}}\)

nhanh vao

\(\sqrt[6]{\left(44-16\sqrt{6}\right)\left(44+16\sqrt{6}\right)}=\sqrt[6]{400}=\sqrt[3]{20}\)

(14,78-a)/(2,87+a)=4/1

14,78+2,87=17,65

Tổng số phần bằng nhau là 4+1=5

Mỗi phần có giá trị bằng 17,65/5=3,53

=>2,87+a=3,53

=>a=0,66.

a, = \(\sqrt[3]{\left(\sqrt{2}+1\right)^3}\) -  \(\sqrt[3]{\left(\sqrt{2}+1\right)^3}\) = \(\sqrt{2}-1-\sqrt{2}-1\)=-2

b, = \(\sqrt{2}+2+2-\sqrt{2}\)=4

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

d, = 

a) Ta có: \(A^3=\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)^3\)

\(=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)

\(=4-3\cdot A\)

\(\Leftrightarrow A^3+3A-4=0\)

\(\Leftrightarrow A^3-A+4A-4=0\)

\(\Leftrightarrow A\left(A-1\right)\left(A+1\right)+4\left(A-1\right)=0\)

\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)

\(\Leftrightarrow A=1\)

a)\(A=^3\sqrt{20+14\sqrt{2}}+^3\sqrt{20-14\sqrt{2}}\)

=>  \(A^3=\left(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\right)^3\)

\(=20+14\sqrt{2}+20-14\sqrt{2}\)

\(+3\left(\text{​​}^3\sqrt{20+14\sqrt{2}}+^3\sqrt{20-14\sqrt{2}}\right)\left(^3\sqrt{20+14\sqrt{2}}.^3\sqrt{20-14\sqrt{2}}\right)\)

\(=40+3A.^3\sqrt{\left(20+14\sqrt{2}\right)\left(20+14\sqrt{2}\right)}\)

\(\Rightarrow A^3=40+3.A.2\)

=> \(A^3-6A-40=0\)

<=> \(A^3-16A+10A-40=0\)

<=> \(A\left(A-4\right)\left(A+4\right)+10\left(A-4\right)=0\)

<=> \(\left(A-4\right)\left(A^2+4A+10\right)=0\)

<=> A = 4 ( vì \(A^2+4A+10=\left(A+2\right)^2+6>0\))

Vậy A = 4.

b/ \(B=^3\sqrt{26+15\sqrt{3}}-^3\sqrt{26-15\sqrt{3}}\)

=> \(B^3=\left(^3\sqrt{26+15\sqrt{3}}-^3\sqrt{26-15\sqrt{3}}\right)^3\)

\(=26+15\sqrt{3}-26+15\sqrt{3}\)

\(-3\left(^3\sqrt{26+15\sqrt{3}}-^3\sqrt{26-15\sqrt{3}}\right).^3\sqrt{26+15\sqrt{3}}.^3\sqrt{26-15\sqrt{3}}\)

\(=30\sqrt{3}-3B.1\)

=> \(B^3+3B-30\sqrt{3}=0\)

<=> \(B^3-12B+15B-30\sqrt{3}=0\)

<=> \(B\left(B-2\sqrt{3}\right)\left(B+2\sqrt{3}\right)+15\left(B-2\sqrt{3}\right)=0\)

<=> \(\left(B-2\sqrt{3}\right)\left(B^2+2\sqrt{3}B+15\right)=0\)

<=> \(B-2\sqrt{3}=0\)( vì \(B^2+2\sqrt{3}B+15=\left(B+\sqrt{3}\right)^2+12>0\))

<=> \(B=2\sqrt{3}\)

e , \(\sqrt{11^2-\left(6\sqrt{2}\right)^2}\)

g, h. Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath