4.tính giá trị biểu thức:
\(A=\sqrt[3]{6\sqrt{3}+10}-\sqrt[3]{6\sqrt{3}-10}\)
\(B=\sqrt[3]{45+29\sqrt{2}}-\sqrt[3]{45-29\sqrt{2}}\)
Tính giá trị biểu thức
A = \(\sqrt[3]{6\sqrt{3}+10}\)TRỪ \(\sqrt[3]{6\sqrt{3}-10}\)
B= \(\sqrt[3]{45+29\sqrt{2}}\) + \(\sqrt[3]{45-29\sqrt{2}}\)
C=\(\sqrt[3]{2+10\sqrt{\frac{1}{27}}}\)+ \(\sqrt[3]{2-10\sqrt{\frac{1}{27}}}\)
A = \(\sqrt[3]{10+6\sqrt{3}}+\sqrt[3]{10-6\sqrt{3}}\)
<=> A3 = 20 - 3×2A
<=> A3 + 6A - 20 = 0
<=> A = 2
Tính giá trị của biểu thức
a. \(A=\sqrt[3]{6\sqrt{3}+10}-\sqrt[3]{6\sqrt{3}-10}\)
b. \(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
c. \(\sqrt[3]{45+29\sqrt{2}}+\sqrt[3]{45-29\sqrt{2}}\)
cảm ơn các bạn trước nhaa
ta có: A3=\(6\sqrt{3}+10-6\sqrt{3}+10-3\sqrt[3]{\left(6\sqrt{3}+10\right)\left(6\sqrt{3}-10\right)}.\left(\sqrt[3]{6\sqrt{3}+10}-\sqrt[3]{6\sqrt{3}-10}\right)\)
=\(20-3.\sqrt[3]{8}.A\)=\(20-6A\)
do đó A3=20-6A↔A3+6A-20=0↔(A2+2A+10)(A-2)=0
dễ thấy A2+2A+10>0→A=2
b) giống a)
c)giống b)
Các bạn cho mình cả cách giải nha!
Thanks các bạn nhìu!!!!!
Bài 1: Giải phương trình sau
a, \(\sqrt[3]{x+1}+\sqrt[3]{7-x}=2\)
b, \(\sqrt[3]{x+3}-\sqrt[3]{6-x}=1\)
Bài 2: Tính giá trị của các biểu thức sau
A=\(\sqrt[3]{6\sqrt{3}+10}-\sqrt[3]{6\sqrt{3}-10}\)
B= \(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
C= \(\sqrt[3]{45+29\sqrt{2}}+\sqrt[3]{45-29\sqrt{2}}\)
D= \(\sqrt[3]{2+10\sqrt{\frac{1}{27}}}+\sqrt[3]{2-10\sqrt{\frac{1}{27}}}\)
1)
dat \(a=\sqrt[3]{x+1};b=\sqrt[3]{7-x}\)
ta co b=2-a
a^3+b^3=x+1+7-x=8
a^3+b^3=a^3+b^3+3ab(a+b)
ab(a+b)=0
suy ra a=0 hoac b=0 hoac a=-b
<=> x=-1; x=7
a=-b
a^3=-b^3
x+1=x+7 (vo li nen vo nghiem)
cau B tuong tu
2)
tat ca cac bai tap deu chung 1 dang do la
\(\sqrt[3]{a+m}+\sqrt[3]{b-m}\)voi m la tham so
dang nay co 2 cach
C1 lap phuong VD: \(B^3=10+3\sqrt[3]{< 5+2\sqrt{13}>< 5-2\sqrt{13}>}\left(B\right)\)
B^3=10-9B
B=1 cach nay nhanh nhung kho nhin
C2 dat an
\(a=\sqrt[3]{5+2\sqrt{13}};b=\sqrt[3]{5-2\sqrt{13}}\)
de thay B=a+b
a^3+b^3=10
ab=-3
B^3=10-9B
suy ra B=1
tuong tu giai cac cau con lai.
Bài 1:
a. Đặt \(a=\sqrt[3]{x+1}\); \(b=\sqrt[3]{7-x}\). Ta có:
\(\hept{\begin{cases}a+b=2\\a^3+b^3=8\end{cases}\Leftrightarrow a^3+\left(2-a\right)^3=8\Leftrightarrow...\Leftrightarrow\orbr{\begin{cases}a=0\\a=2\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}a=0\\b=2\end{cases}}\)hoặc \(\hept{\begin{cases}a=2\\b=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}\sqrt[3]{x+1}=0\\\sqrt[3]{7-x}=2\end{cases}}\)hoặc \(\hept{\begin{cases}\sqrt[3]{x+1}=2\\\sqrt[3]{7-x}=0\end{cases}}\)
\(\Leftrightarrow x=-1\)hoặc \(x=7\)
Bài 2:
\(A=\sqrt[3]{\left(\sqrt{3}\right)^3+3.\left(\sqrt{3}\right)^2.1+3.\sqrt{3}.1^2+1^3}-\sqrt[3]{\left(\sqrt{3}\right)^3-3.\left(\sqrt{3}\right)^2.1+3.\sqrt{3}.1^2-1^3}\)
\(=\sqrt[3]{\left(\sqrt{3}+1\right)^3}-\sqrt[3]{\left(\sqrt{3}-1\right)^3}=\sqrt{3}+1-\sqrt{3}+1=2\)
Những câu khác làm tương tự.
2.so sánh
\(a.\sqrt[3]{5\sqrt{2}-7}-33\sqrt{2}và-1\)
3.tính giá trị của biểu thức:
\(B=\sqrt[3]{45+29\sqrt{2}}-\sqrt[3]{45-29\sqrt{2}}\)
Tính
1, a = \(\sqrt[3]{45+26\sqrt{2}}+\sqrt[3]{45-29\sqrt{2}}\)
2, x = \(\sqrt[3]{4+\sqrt{80}-\sqrt[3]{\sqrt{80}-4}}\)
3, \(\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
4, \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
5, \(\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}+\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}\)
Rút gọn các biểu thức sau :
a,\(\dfrac{\sqrt{6}+\sqrt{10}}{\sqrt{21}+\sqrt{35}}\)
b,\(\dfrac{\sqrt{405}+3\sqrt{27}}{3\sqrt{3}+\sqrt{45}}\)
c,\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6}-\sqrt{9}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
d, D=\(\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{9\left(x^2+2xy+y^2\right)}{4}}\) \(\left(vớix\ne y,x\ne-y\right)\)
d: \(D=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{9\left(x^2+2xy+y^2\right)}{4}}\)
\(=\dfrac{2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{3\left(x+y\right)}{2}\)
\(=\dfrac{3}{x-y}\)
Tính
1, a = \(\sqrt[3]{45+29\sqrt{2}}+\sqrt[3]{45-29\sqrt{2}}\)
2, x = \(\sqrt[3]{4+\sqrt{80}-\sqrt[3]{\sqrt{80}-4}}\)
3, \(\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
4, \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
5,\(\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}+\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}\)
3: \(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
4: \(=\dfrac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}=-\sqrt{2}\)
5: \(=\dfrac{\sqrt{23-8\sqrt{7}}}{3}+\dfrac{\sqrt{23+8\sqrt{7}}}{3}\)
\(=\dfrac{4-\sqrt{7}+4+\sqrt{7}}{3}=\dfrac{8}{3}\)
Tính giá trị biểu thức
C= \(\sqrt[3]{45+29\sqrt{2}}+\sqrt[3]{45-29\sqrt{2}}\)
C= 3√45+29√2+3√45−29√2
⇔\(C^3=45+29\sqrt{2}+45-29\sqrt{2}+3\sqrt[3]{45+29\sqrt{2}}.\sqrt[3]{45-29\sqrt{2}}\left(\sqrt[3]{45+29\sqrt{2}}+\sqrt[3]{45-29\sqrt{2}}\right)\\ C^3=90+3\sqrt[3]{343}.C\\ C^3=90+21C\\ C^3-21C-90=0\\ C^3-36C+15C-90\\ C\left(C-6\right)\left(C+6\right)+15\left(C-6\right)=0\\ \left(C-6\right)\left[C\left(C+6\right)+15\right]=0\\ \left(C-6\right)\left(C^2+6C+15\right)=0\\ \)
Mà C2+6C+15=(C+3)2+6 > 0
Nên C-6=0
⇒C=6
Tính giá trị biểu thức:
a) \(\frac{\sqrt[3]{4}+\sqrt[3]{2}+2}{\sqrt[3]{4}+\sqrt[3]{2}+1}\)
b) \(\sqrt{3+\sqrt{3}+\sqrt[3]{10+6\sqrt{3}}}\)
c) \(\frac{4+2\sqrt{3}}{\sqrt[3]{10+6\sqrt{3}}}\)