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VD
2
2 tháng 8 2019
5:x^2 +4x +5x + 20 =0
(x^2 + 4x).(5x+20)
x(x+4).5(x+4)
(x+4).(x+5)
[x+5=0 ->x=-5
[x+4=0 ->x=-4
18 tháng 6 2017
\(\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-166}{23}=10\)
\(\Leftrightarrow\left(\frac{x-241}{17}-1\right)+\left(\frac{x-220}{19}-2\right)+\left(\frac{x-195}{21}-3\right)+\left(\frac{x-166}{23}-4\right)=0\)
\(\Leftrightarrow\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{21}+\frac{x-258}{23}=0\)
\(\Leftrightarrow\left(x-258\right)\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23}\right)=0\)
\(\Leftrightarrow x-258=0\)(vì \(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23}\ne0\))
\(\Leftrightarrow x=258\)
vậy phương trình có tập nghiệm là: S={258}
25 tháng 10 2020
Bài làm
Rút gọn
\(\left(\frac{1}{\sqrt{x}-1}-\frac{\sqrt{x}}{1-x}\right)\cdot\frac{x-\sqrt{x}}{2\sqrt{x}+1}\)
\(=\left(\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right)\cdot\frac{\sqrt{x}(\sqrt{x}-1)}{2\sqrt{x}+1}\)
\(=\left(\frac{\sqrt{x}+1}{(\sqrt{x}-1)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\frac{\sqrt{x}(\sqrt{x}-1)}{2\sqrt{x}+1}\)
\(=\frac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}+1}\)
\(=\frac{\sqrt{x}}{\sqrt{x}+1}\)
Tính:
\(\frac{3-\sqrt{3}}{\sqrt{3}+2}+\frac{\sqrt{3}}{\sqrt{3}-2}+\frac{21}{\sqrt{3}}\)
\(=\frac{3-\sqrt{3}}{\sqrt{3}+2}+\frac{\sqrt{3}}{\sqrt{3}-2}+\frac{7\sqrt{3}\cdot\sqrt{3}}{\sqrt{3}}\)
\(=\frac{3-\sqrt{3}}{\sqrt{3}+2}+\frac{\sqrt{3}}{\sqrt{3}-2}+7\sqrt{3}\)
\(=\frac{\left(3-\sqrt{3}\right)\left(\sqrt{3}-2\right)}{\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)}+\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}+7\sqrt{3}\)
\(=\frac{3\sqrt{3}-3-6+2\sqrt{3}}{\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)}+\frac{3+2\sqrt{3}}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}+7\sqrt{3}\)
\(=\frac{3\sqrt{3}-3-6+2\sqrt{3}+3+2\sqrt{3}}{3-4}+7\sqrt{3}\)
\(=\frac{7\sqrt{3}-6}{-1}+7\sqrt{3}\)
\(=6-7\sqrt{3}+7\sqrt{3}\)
\(=6\)
25 tháng 10 2020
Bài làm
\(\sqrt{42-10\sqrt{17}}+\sqrt{\left(\sqrt{17}-\sqrt{16}\right)^2}\)
\(=\sqrt{42-10\sqrt{17}}+\left|\sqrt{17}-\sqrt{16}\right|\)
\(=\sqrt{25-10\sqrt{17}+17}+\sqrt{17}-\sqrt{16}\)
\(=\sqrt{\left(5-\sqrt{17}\right)^2}+\sqrt{17}-\sqrt{16}\)
\(=\left|5-\sqrt{17}\right|+\sqrt{17}-\sqrt{16}\)
\(=5-\sqrt{17}+\sqrt{17}-\sqrt{16}\)
\(=5-4\)
\(=1\)
9 tháng 7 2019
a, \(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)
\(\Rightarrow\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9\left(x-1\right)}+24\frac{\sqrt{x-1}}{\sqrt{64}}=-17\)
\(\Rightarrow\frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+\frac{24\sqrt{x-1}}{8}=-17\)
\(\Rightarrow\frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Rightarrow\sqrt{x-1}\left(\frac{1}{2}-\frac{9}{2}+3\right)=-17\)
\(\Rightarrow\sqrt{x-1}.-1=-17\)
\(\Rightarrow\sqrt{x-1}=17\)
\(\Rightarrow x-1=289\)
\(\Rightarrow x=290\)
b, \(3x-7\sqrt{x}+4=0\)
\(\Rightarrow3x-3\sqrt{x}-4\sqrt{x}+4=0\)
\(\Rightarrow3\sqrt{x}\left(\sqrt{x}-1\right)-4\left(\sqrt{x}-1\right)=0\)
\(\Rightarrow\left(\sqrt{x}-1\right)\left(3\sqrt{x}-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}-1=0\\3\sqrt{x}-4=0\end{cases}\Rightarrow}\orbr{\begin{cases}\sqrt{x}=1\\3\sqrt{x}=4\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{16}{9}\end{cases}}}\)
c, \(-5x+7\sqrt{x}+12=0\)
\(\Rightarrow-5x-5\sqrt{x}+12\sqrt{x}+12=0\)
\(\Rightarrow-5\sqrt{x}\left(\sqrt{x}+1\right)+12\left(x+1\right)=0\)
\(\Rightarrow\left(\sqrt{x}+1\right)\left(-5\sqrt{x}+12\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}+1=0\\-5\sqrt{x}+12=0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}=-1VN\\-5\sqrt{x}=-12\end{cases}}\Rightarrow\orbr{\begin{cases}\\\sqrt{x}=\frac{12}{5}\end{cases}\Rightarrow}\orbr{\begin{cases}\\x=\frac{144}{25}\end{cases}}}\)
9 tháng 7 2019
1) ĐK: \(x-1\ge0\Leftrightarrow x\ge1\)
pt \(\Leftrightarrow\frac{1}{2}\sqrt{x-1}-\frac{3}{2}.3\sqrt{x-1}+\frac{24}{8}\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}\left(\frac{1}{2}-\frac{9}{2}+3\right)=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=17^2=289\Leftrightarrow x=290\left(tm\right)\)
b) \(3x-7\sqrt{x}+4=0\)
ĐK: \(x\ge0\)
Đặt \(\sqrt{x}=t\left(t\ge0\right)\Leftrightarrow t^2=x\)
Ta có phương trình ẩn t:
\(3t^2-7t+4=0\)( giải đen ta)
\(\Leftrightarrow\orbr{\begin{cases}t=1\\t=\frac{4}{3}\end{cases}}\)
Với t=1 ta có: \(\sqrt{x}=1\Leftrightarrow x=1\) (tm)
Với t=4/3 ta có: \(\sqrt{x}=\frac{4}{3}\Leftrightarrow x=\frac{16}{9}\) (tm)
Câu c em làm tương tự câu b nhé!
MN
3 tháng 3 2020
\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne1\\x\ne9\end{cases}}\)
\(A=\sqrt{x}+1-\frac{17}{1-\sqrt{x}}\)
\(\Leftrightarrow A=\frac{x-1+17}{\sqrt{x}-1}\)
\(\Leftrightarrow A=\frac{x+16}{\sqrt{x}-1}\)
\(B=\frac{x-7}{x-4\sqrt{x}+3}+\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}-3}\)
\(\Leftrightarrow B=\frac{x-7+\sqrt{x}-3-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow B=\frac{x-9}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow B=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow B=\frac{\sqrt{x}+3}{\sqrt{x}-1}\)
Vậy \(P=\frac{x-16}{\sqrt{x}-1}:\frac{\sqrt{x}+3}{\sqrt{x}-1}\)
\(\Leftrightarrow P=\frac{\left(x-16\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow P=\frac{x-16}{\sqrt{x}+3}\)
b) Ta có : \(\sqrt{x}+3\ge3>0\)
Để P min \(\Leftrightarrow x-16\) min
Mà \(x-16\ge-16\)
Dấu " = " xảy ra \(\Leftrightarrow x=0\)
Vậy \(Min_P=\frac{-16}{3}\Leftrightarrow x=0\)
x=101 nha.
x=101
mk chắc chắn
:v