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Ta có: \(2014a=2016-3a-a^3\)
\(S=\sqrt[3]{3a^2+2014-2015}+\sqrt[3]{3a^2-2014a+2017}\)
\(=\sqrt[3]{3a^2+2016-3a-a^3-2015}+\sqrt[3]{3a^2-2016+3a+a^3+2017}\)
\(=\sqrt[3]{-a^3+3a^2-3a+1}+\sqrt[3]{a^3+3a^2+3a+1}\)
\(=\sqrt[3]{\left(1-a\right)^3}+\sqrt[3]{\left(1+a\right)^3}=1-a+1+a=2\)
Dat bieu thuc tren la A
ta co \(\frac{1}{\sqrt{n+2}+\sqrt{n}}=\frac{\sqrt{n+2}-\sqrt{n}}{2}\)
ap dung dang thuc tren ta co\(\frac{1}{\sqrt{3}+1}=\frac{\sqrt{3}-1}{2}\)
tuong tu ta co \(\frac{1}{\sqrt{5}+\sqrt{3}}=\frac{\sqrt{5}-\sqrt{3}}{2}\)
.........
\(\frac{1}{\sqrt{2017}+\sqrt{2015}}=\frac{\sqrt{2017}-\sqrt{2015}}{2}\)
ta co
\(A=\frac{1}{2}\left(\sqrt{3}-1+\sqrt{5}-\sqrt{3}+.....+\sqrt{2017}-\sqrt{2015}\right)=\frac{\sqrt{2017}-1}{2}\)
\(x=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
\(\Rightarrow x^3=5+2\sqrt{13}+5-2\sqrt{13}+3\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}.x\)
\(=10+3x\sqrt[3]{25-52}\)
\(=10+3x\sqrt[3]{-27}\)
\(=10-9x\)
\(\Rightarrow x^3+9x-10=0\)
\(\Leftrightarrow x^3-x+10x-10=0\)
\(\Leftrightarrow x\left(x^2-1\right)+10\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)+10\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+10\right)=0\)
Vì \(x^2+x+10=\left(x+\frac{1}{2}\right)^2+\frac{39}{4}>0\forall x\)
=> x - 1 = 0
=> x = 1
Thay vào A = 12015 - 12016 = 0
Vậy A = 0
\(\text{Ta có: }x=\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}}=\sqrt{\frac{\left(3-\sqrt{5}\right)^2}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}}=\frac{3-\sqrt{5}}{\sqrt{9-5}}=\frac{3-\sqrt{5}}{2}.\)
\(A=x^5-6x^4+12x^3-4x^2-13x+2020\)
\(=\left(x^5-3x^4+x^3\right)-\left(3x^4-9x^3+3x^2\right)+\left(2x^3-6x^2+2x\right)+\left(5x^2-15x+5\right)+2015\)
\(=x^3\left(x^2-3x+1\right)-3x^2\left(x^2-3x+1\right)+2x\left(x^2-3x+1\right)+5\left(x^2-3x+1\right)+2015\)
\(=\left(x^2-3x+1\right)\left(x^3-3x^2+2x+5\right)+2015\)
Thay x vào A ta có:
\(A=\left[\left(\frac{3-\sqrt{5}}{2}\right)^2-3.\frac{3-\sqrt{5}}{2}+1\right]\left(.....\right)+2015\)
\(=\left(\frac{14-6\sqrt{5}}{4}-\frac{9-3\sqrt{5}}{2}+1\right)\left(....\right)+2015\)
\(=0\cdot\left(......\right)+2015=2015\)
Vậy.....
x+√(x^2+3)=3/(y+√(y^3))=3(y-√(y^2+3)/-a(trục căn thức)
x+√(x^2+3)=-y+√(y^2+3) suy ra x+y=√(y^2+3)-√(x^2+3)(1)
Tương tự,x+y=√(x^2+3)-√(y^2+3)(2)
Cộng (1),(2) theo vế suy ra 2(x+y)=0 suy ra x+y=0
hay E=0.
Vậy E=0
nhân \(-x+\sqrt{x^2+3}\) vào 2 vế ta đc : \(\left(-x^2+x^2+3\right)\left(y+\sqrt{y^2+3}\right)=\)\(3\left(-x+\sqrt{x^2+3}\right)\)
<=> \(y+\sqrt{y^2+3}=-x+\sqrt{x^2+3}\)<=> \(y+\sqrt{y^2+3}+x-\sqrt{x^2+3}=0\)__(1)___
làm tương tự ta đc \(\left(-y+\sqrt{y^2+3}\right)\left(x+\sqrt{x^2+3}\right)\)\(=3\left(-y+\sqrt{y^2+3}\right)\)
<=> \(x+\sqrt{x^2+3}=-y+\sqrt{y^2+3}\)<=> \(x+\sqrt{x^2+3}+y-\sqrt{y^2+3}=0\)__(2)__
lấy (1) + (2) => 2(x+y) =0 => x+y=0
lấy
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 \(x=7\)
ai nay dung kinh nghiem la chinh
cau a)
ta thay \(10+6\sqrt{3}=\left(1+\sqrt{3}\right)^3\)
\(6+2\sqrt{5}=\left(1+\sqrt{5}\right)^2\)
khi do \(x=\frac{\sqrt[3]{\left(\sqrt{3}+1\right)^3}\left(\sqrt{3}-1\right)}{\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{5}}\)
\(x=\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{1+\sqrt{5}-\sqrt{5}}\)
\(x=\frac{3-1}{1}=2\)
suy ra
x^3-4x+1=1
A=1^2018
A=1
b)
ta thay
\(7+5\sqrt{2}=\left(1+\sqrt{2}\right)^3\)
khi do
\(x=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\frac{1}{\sqrt[3]{\left(1+\sqrt{2}\right)^3}}\)
\(x=1+\sqrt{2}-\frac{1}{1+\sqrt{2}}=\frac{\left(1+\sqrt{2}\right)^2-1}{1+\sqrt{2}}=\frac{2+2\sqrt{2}}{1+\sqrt{2}}\)
x=2
thay vao
x^3+3x-14=0
B=0^2018
B=0
CTV ơi giải dùm mình với.
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