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\(x=\frac{\left(\sqrt{3}-1\right)\sqrt[3]{10+6\sqrt{3}}}{\sqrt{21+4\sqrt{5}}+3}=\frac{\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+1\right)}{2\sqrt{5}+1+3}=\frac{2}{2\sqrt{5}+4}=\sqrt{5}-2\)
\(A=\left(x^2-4x-2\right)^{2015}\)
Xong chắc thay vào nhưng để nguyên số mũ ko tính ra
1) \(\sqrt[]{9\left(x-1\right)}=21\)
\(\Leftrightarrow9\left(x-1\right)=21^2\)
\(\Leftrightarrow9\left(x-1\right)=441\)
\(\Leftrightarrow x-1=49\Leftrightarrow x=50\)
2) \(\sqrt[]{1-x}+\sqrt[]{4-4x}-\dfrac{1}{3}\sqrt[]{16-16x}+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}+\sqrt[]{4\left(1-x\right)}-\dfrac{1}{3}\sqrt[]{16\left(1-x\right)}+5=0\)
\(\)\(\Leftrightarrow\sqrt[]{1-x}+2\sqrt[]{1-x}-\dfrac{4}{3}\sqrt[]{1-x}+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}\left(1+3-\dfrac{4}{3}\right)+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}.\dfrac{8}{3}=-5\)
\(\Leftrightarrow\sqrt[]{1-x}=-\dfrac{15}{8}\)
mà \(\sqrt[]{1-x}\ge0\)
\(\Leftrightarrow pt.vô.nghiệm\)
3) \(\sqrt[]{2x}-\sqrt[]{50}=0\)
\(\Leftrightarrow\sqrt[]{2x}=\sqrt[]{50}\)
\(\Leftrightarrow2x=50\Leftrightarrow x=25\)
1) \(\sqrt{9\left(x-1\right)}=21\) (ĐK: \(x\ge1\))
\(\Leftrightarrow3\sqrt{x-1}=21\)
\(\Leftrightarrow\sqrt{x-1}=7\)
\(\Leftrightarrow x-1=49\)
\(\Leftrightarrow x=49+1\)
\(\Leftrightarrow x=50\left(tm\right)\)
2) \(\sqrt{1-x}+\sqrt{4-4x}-\dfrac{1}{3}\sqrt{16-16x}+5=0\) (ĐK: \(x\le1\))
\(\Leftrightarrow\sqrt{1-x}+2\sqrt{1-x}-\dfrac{4}{3}\sqrt{1-x}+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}=-5\) (vô lý)
Phương trình vô nghiệm
3) \(\sqrt{2x}-\sqrt{50}=0\) (ĐK: \(x\ge0\))
\(\Leftrightarrow\sqrt{2x}=\sqrt{50}\)
\(\Leftrightarrow2x=50\)
\(\Leftrightarrow x=\dfrac{50}{2}\)
\(\Leftrightarrow x=25\left(tm\right)\)
4) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\left(ĐK:x\ge-\dfrac{1}{2}\right)\\2x+1=-6\left(ĐK:x< -\dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(tm\right)\\x=-\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)
5) \(\sqrt{\left(x-3\right)^2}=3-x\)
\(\Leftrightarrow\left|x-3\right|=3-x\)
\(\Leftrightarrow x-3=3-x\)
\(\Leftrightarrow x+x=3+3\)
\(\Leftrightarrow x=\dfrac{6}{2}\)
\(\Leftrightarrow x=3\)
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
\(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}+\frac{5\left(2\sqrt{2}+\sqrt{3}\right)}{\left(2\sqrt{2}+\sqrt{3}\right)\left(2\sqrt{2}-\sqrt{3}\right)}-\frac{5\left(\sqrt{8}-\sqrt{3}\right)}{\left(\sqrt{8}-\sqrt{3}\right)\left(\sqrt{8}+\sqrt{3}\right)}\)
\(=\sqrt{3}+1+\sqrt{3}-1+\frac{5\left(2\sqrt{2}+\sqrt{3}\right)}{5}-\frac{5\left(\sqrt{8}-\sqrt{3}\right)}{5}\)
\(=2\sqrt{3}+2\sqrt{2}+\sqrt{3}-\sqrt{8}+\sqrt{3}\)
\(=4\sqrt{3}\)
Giải pt:
1/ \(\Leftrightarrow2x-1=5\)
\(\Leftrightarrow2x=6\Rightarrow x=3\)
2/ \(\Leftrightarrow\sqrt{3}x^2=\sqrt{12}\Leftrightarrow x^2=\sqrt{4}=2\)
\(\Leftrightarrow x=\pm\sqrt{2}\)
3/ \(\Leftrightarrow x-5=9\Rightarrow x=14\)
4/ Đề thiếu
5/ \(\Leftrightarrow\left|x-3\right|=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=9\\x-3=-9\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-6\end{matrix}\right.\)
6/ \(\Leftrightarrow2\left|1-x\right|=6\)
\(\Leftrightarrow\left|1-x\right|=3\Leftrightarrow\left[{}\begin{matrix}1-x=3\\1-x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\)
7/ \(\Leftrightarrow9\left(x-1\right)=21^2\)
\(\Leftrightarrow x-1=49\Rightarrow x=50\)
8/ \(\Leftrightarrow x+1=2^3=8\)
\(\Rightarrow x=7\)
9/ \(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=-\frac{7}{2}\end{matrix}\right.\)
10/ \(\Leftrightarrow\sqrt{2}x=\sqrt{50}\Leftrightarrow x=\sqrt{25}\Rightarrow x=5\)
11/ \(\Leftrightarrow\left|2x-1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
12/ \(\Leftrightarrow3-2x=\left(-2\right)^3=-8\)
\(\Leftrightarrow2x=11\Rightarrow x=\frac{11}{2}\)
bài 1:
a)\(\left(3-\sqrt{2}\right)\sqrt{7+4\sqrt{3}}\)
\(=\left(3-\sqrt{2}\right)\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\left(3-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)\(do2>\sqrt{3}\)
\(=6+3\sqrt{3}-2\sqrt{2}-\sqrt{6}\)
b) \(\left(\sqrt{3}+\sqrt{5}\right)\sqrt{7-2\sqrt{10}}\)
\(=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\)
\(=\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)do\sqrt{5}>\sqrt{2}\)
\(=\sqrt{15}-\sqrt{6}+5-\sqrt{10}\)
c)\(\left(2+\sqrt{5}\right)\sqrt{9-4\sqrt{5}}\)
\(=\left(2+\sqrt{5}\right)\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(=\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)do\sqrt{5}>2\)
\(=5-4\)
\(=1\left(hđt.3\right)\)
d)\(\left(\sqrt{6}+\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)
\(=\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{8-2\sqrt{15}}\)
\(=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)do\sqrt{5}>\sqrt{3}\)
\(=5-3\)
\(=2\)
e)\(\sqrt{2}\left(\sqrt{8}-\sqrt{32}+3\sqrt{18}\right)\)
\(=\sqrt{2}\left(2\sqrt{2}-4\sqrt{2}+9\sqrt{2}\right)\)
\(=2\left(2-4+9\right)\)
\(=2.7=14\)
f)\(\sqrt{2}\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)\)
\(=2-\sqrt{6-2\sqrt{5}}\)
\(=2-\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=2-\left(\sqrt{5}-1\right)\)
\(=2-\sqrt{5}+1\)
\(=3-\sqrt{5}\)
g)\(\sqrt{3}-\sqrt{2}\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=\sqrt{3}-\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)\)
\(=\sqrt{3}-\sqrt{6}-2\)
h) \(\left(\sqrt{2}-\sqrt{3+\sqrt{5}}\right)\sqrt{2}+2\sqrt{5}\)
\(=\left(2-\sqrt{6+2\sqrt{5}}\right)+2\sqrt{5}\)
\(=\left(2-\sqrt{\left(\sqrt{5}+1\right)^2}\right)+2\sqrt{5}\)
\(=2-\left(\sqrt{5}+1\right)+2\sqrt{5}\left(do\sqrt{5}>1\right)\)
\(=2-\sqrt{5}-1+2\sqrt{5}\)
\(=1-\sqrt{5}\)
bài 2)
a) \(\sqrt{4x^2-4x+1}=5\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=5\)
\(\Leftrightarrow2x-1=5\)hoặc \(\Leftrightarrow2x-1=-5\)
\(\Leftrightarrow x=3\)hoặc \(\Leftrightarrow x=-2\)
Vậy x = 3 hoặc x = -2