Cho x=\(\dfrac{1}{\sqrt[3]{4-\sqrt{15}}}+\sqrt[3]{4-\sqrt{15}}\)
Tính P=x3-3x+2006
tìm ĐKXĐ
1, \(\sqrt{6x+1}\)
2,\(\dfrac{\sqrt{3}-4}{\sqrt{3x-5}}\)
3, \(\sqrt{\dfrac{2\sqrt{15}-\sqrt{59}}{x-7}}\)
4,\(\sqrt{\dfrac{-3x}{1-\sqrt{2}}}\)
5, \(\sqrt{\sqrt{5}-\sqrt{3}x}\)
1.
6x + 1 ≥0
<=>6x≥-1
<=>x≥-1/6
2.
3x - 5 > 0
<=> 3x > 5
<=> x > 5/3
5.
√5 - √3 . x ≥0
<=> √3 . x ≤ √5
<=> x ≤ √5/3 = (√15)/3
Cho x=\(\frac{1}{\sqrt[3]{4-\sqrt{15}}}+\sqrt[3]{4-\sqrt{15}}\). Tính A= \(x^3\)+ 3x + 2006
Các bạn trả lời dùm mình nhak, cần gấp lắm. Bạn nào nhanh mình tick cho :))
\(x=\frac{1}{\sqrt[3]{4-\sqrt{15}}}+\sqrt[3]{4-\sqrt{15}}\)
<=> \(x^3=\frac{1}{4-\sqrt{15}}+3\left(\frac{1}{\sqrt[3]{4-\sqrt{15}}}+\sqrt[3]{4-\sqrt{15}}\right)\left(\frac{1}{\sqrt[3]{4-\sqrt{15}}}.\sqrt[3]{4-\sqrt{15}}\right)\)
\(+4-\sqrt{15}\)
<=> \(x^3=\frac{1}{4-\sqrt{15}}+4-\sqrt{15}+3x\)
<=> \(x^3-3x+2006=\frac{1}{4-\sqrt{15}}+4-\sqrt{15}+2006\)
<=> \(x^3-3x+2006=\frac{4+\sqrt{15}}{16-15}+4-\sqrt{15}+2006\)
<=> \(x^3-3x+2006=2014\)
Cho \(x=\dfrac{\sqrt[3]{4+\sqrt{15}}+\sqrt[3]{4-\sqrt{15}}+1}{2}\)
Tính \(P=12x^5+18x^4+4x^3-15x^2-21\)
chắc bạn chép sai đề rồi , hai căn đầu phải 1 cộng 1 trừ chứ
Đặt
\(x=\dfrac{y+1}{2}\Rightarrow y=2x-1\)
\(\Rightarrow y=\sqrt[3]{4+\sqrt{15}}+\sqrt[3]{4-\sqrt{15}}\)
\(y^3=8+3\sqrt[3]{\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)}=8+3y\)
\(\Rightarrow y^3-3y-8=0\\ \)
\(\Leftrightarrow8x^3-12x^2-6=0\)
\(\Rightarrow4x^3-6x^2-3=0\)
thay p vào ta có
\(P=12x^5-18x^4+4x^3-15x^2-21\)
\(=12x^5-18x^4-9x^2-4x^3-6x^2-21\)
\(=3x^2\left(4x^2-6x^2-3\right)+4x^3-6x^2-3\\ =3x^2.0+0-18\\ =-18\)
Tìm x:
a)\(\dfrac{1}{3}\sqrt{x-1}+2\sqrt{4x-4}-12\sqrt{\dfrac{x-1}{25}}=\dfrac{29}{15}\)
b)\(\dfrac{3x-2}{\sqrt{x-1}}-\sqrt{x+1}=\sqrt{2x-3}\)
Cho \(x=\frac{1}{\sqrt[3]{4-\sqrt{15}}}+\sqrt[3]{4-\sqrt{15}}\)
Tính \(y=x^3-3x+1987\)
\(x=\frac{1}{\sqrt[3]{4-\sqrt{15}}}+\sqrt[3]{4-\sqrt{15}}\)
\(\Leftrightarrow x^3=\frac{1}{4-\sqrt{15}}+4-\sqrt{15}+3\sqrt[3]{\sqrt[3]{\frac{1}{4-\sqrt{5}}}.\sqrt[3]{4-\sqrt{5}}}.x\)
\(=4+\sqrt{15}+4-\sqrt{15}+3x=8+3x\)
=>y=3x+8-3x+1987
=1995
Cho \(x=\dfrac{\sqrt{2}-1}{1+2}+\dfrac{\sqrt{3}-\sqrt{2}}{2+3}+\dfrac{\sqrt{4}-\sqrt{3}}{3+4}+...+\dfrac{\sqrt{225}-\sqrt{224}}{224+225}\) . Chứng minh rằng \(x< \dfrac{7}{15}\) .
Thực hiện từng bước của phép tính:
1.\(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3\)
2.\(\sqrt{4-\sqrt{15}}+\sqrt{4+\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
3.\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)
4.\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
1.\(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3=2\sqrt{2}+6+3\sqrt{2}+1-\left(2\sqrt{2}-6+3\sqrt{2}-1\right)=14\)
2.\(\sqrt{4-\sqrt{15}}+\sqrt{4+\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
\(=\sqrt{\dfrac{1}{2}\left(8-2\sqrt{3.}\sqrt{5}\right)}+\sqrt{\dfrac{1}{2}\left(8+2.\sqrt{3}.\sqrt{5}\right)}-\sqrt{2}\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\dfrac{1}{2}\left(\sqrt{3}-\sqrt{5}\right)^2}+\sqrt{\dfrac{1}{2}\left(\sqrt{3}+\sqrt{5}\right)^2}-\sqrt{2}\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\dfrac{\sqrt{2}}{2}\left|\sqrt{3}-\sqrt{5}\right|+\dfrac{\sqrt{2}}{2}\left(\sqrt{3}+\sqrt{5}\right)-\sqrt{2}\left|\sqrt{5}-1\right|\)
\(=\dfrac{\sqrt{2}}{2}\left(\sqrt{5}-\sqrt{3}\right)+\dfrac{\sqrt{2}}{2}\left(\sqrt{3}+\sqrt{5}\right)-\sqrt{2}\left(\sqrt{5}-1\right)\)
\(=\sqrt{5}.\sqrt{2}-\sqrt{2}\left(\sqrt{5}-1\right)=\sqrt{2}\)
3.\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}=\dfrac{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8\left(1+\sqrt{5}\right)}{1-\left(\sqrt{5}\right)^2}\)
\(=\sqrt{20}+\dfrac{8\left(1+\sqrt{5}\right)}{-4}=2\sqrt{5}-2\left(1+\sqrt{5}\right)=-2\)
4.\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
\(=\sqrt{\dfrac{4-2\sqrt{3}}{4+2\sqrt{3}}}+\sqrt{\dfrac{4+2\sqrt{3}}{4-2\sqrt{3}}}\)\(=\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{\left(\sqrt{3}+1\right)^2}}+\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}-1\right)^2}}\)
\(=\dfrac{\left|\sqrt{3}-1\right|}{\sqrt{3}+1}+\dfrac{\sqrt{3}+1}{\left|\sqrt{3}-1\right|}=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}+\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)
\(=\dfrac{\left(\sqrt{3}-1\right)^2+\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\dfrac{8}{3-1}=4\)
3: Ta có: \(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)
\(=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-\dfrac{8\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)
\(=2\sqrt{5}-2\left(\sqrt{5}+1\right)\)
=-2
4) Ta có: \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}\)
=4
1.\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
2.\(\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}+1}\)
3.\(\left(1-\dfrac{4}{\sqrt{x}+1}+\dfrac{1}{x-1}\right):\dfrac{x-2\sqrt{x}}{x-1}\)
4.\(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
4 , Ta có :
\(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x-9}{x-9}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}-\dfrac{3\left(x-3\right)}{x-9}\)
\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x+9}{x-9}\)
\(=\dfrac{3\sqrt{x}+9}{x-9}\)
\(=\dfrac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{3}{\sqrt{x}-3}\)
2 , Ta có :
\(\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}+1}=\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{\left(x-1\right)\left(\sqrt{x}-1\right)}{x-1}\)
\(=\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x\sqrt{x}-x-\sqrt{x}+1}{x-1}\)
\(=\dfrac{x\sqrt{x}+1-x\sqrt{x}+x+\sqrt{x}-1}{x-1}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
1: \(=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
3: \(=\dfrac{x-1-4\sqrt{x}+4+1}{x-1}\cdot\dfrac{x-1}{x-2\sqrt{x}}\)
\(=\dfrac{x-4\sqrt{x}+4}{x-2\sqrt{x}}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
Rút gọn biểu thức sau:
Q =\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+x}+\dfrac{2+3x}{4-x}\)
\(Q=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+5\sqrt{x}-8}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(P=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-3x}{x-4}\)
\(=\dfrac{-\sqrt{x}}{x-4}\)