Giaỉ phương trình:
a) \(3x-7\sqrt{x}+4=0\)
b) \(\frac{\sqrt{x-1}}{\sqrt{x+3}}=\frac{\sqrt{x-2}}{\sqrt{x+1}}\)
c) \(x-5\sqrt{x-2}=2\)
d) \(\sqrt{x-2\sqrt{x}+1}=2\)
Các bạn giúp mk nhé!!! ~ Thanks ~
Giaỉ phương trình:
a) \(3x-7\sqrt{x}+4=0\)
b) \(\frac{\sqrt{x-1}}{\sqrt{x+3}}=\frac{\sqrt{x-2}}{\sqrt{x+1}}\)
c) \(x-5\sqrt{x-2}=2\)
d) \(\sqrt{x-2\sqrt{x}+1}=2\)
a ; \(3x-7\sqrt{x}+4=0
\)
\(3x-3\sqrt{x}-4\sqrt{x}+4=0\)\(\left(\sqrt{x}-1\right)\left(3\sqrt{x}-4\right)=0\)
từ đó suy ra x
Bạn giải cụ thể từng câu cho mk nhé!!! :))))
giaỉ các phương trình vô tỉ sau
\(x^2-3x+1+\frac{\sqrt{3}}{3}.\sqrt{x^4+x^2+1}=0\)
\(\sqrt[3]{4+4x-x^2}+x\sqrt{x\left(6-x^2\right)}+3x=12+\sqrt{2-x}\)
Giải các phương trình:
a) \(\sqrt{x^2-3x+2}=\sqrt{x-1}\)
b) \(\sqrt{x^2-4x+4}=\sqrt{4x^2-12x+9}\)
c) \(\sqrt{x^2-5x+6}=\sqrt{x-2}\)
d) \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\)
a. ĐKXĐ: $x\geq 2$ hoặc $x=1$
PT $\Leftrightarrow \sqrt{(x-1)(x-2)}=\sqrt{x-1}$
$\Leftrightarrow \sqrt{x-1}(\sqrt{x-2}-1)=0$
\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-1}=0\\ \sqrt{x-2}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=3\end{matrix}\right.\) (đều thỏa mãn)
b.
PT $\Leftrightarrow \sqrt{(x-2)^2}=\sqrt{(2x-3)^2}$
$\Leftrightarrow |x-2|=|2x-3|$
\(\Leftrightarrow \left[\begin{matrix} x-2=2x-3\\ x-2=3-2x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=\frac{5}{3}\end{matrix}\right.\)
c. ĐKXĐ: $x=2$ hoặc $x\geq 3$
PT $\Leftrightarrow \sqrt{(x-2)(x-3)}=\sqrt{x-2}$
$\Leftrightarrow \sqrt{x-2}(\sqrt{x-3}-1)=0$
\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-2}=0\\ \sqrt{x-3}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=2\\ x=4\end{matrix}\right.\) (đều tm)
d.
PT $\Leftrightarrow \sqrt{(2x-1)^2}=\sqrt{(x-3)^2}$
$\Leftrightarrow |2x-1|=|x-3|$
\(\Leftrightarrow \left[\begin{matrix} 2x-1=x-3\\ 2x-1=3-x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=-2\\ x=\frac{4}{3}\end{matrix}\right.\)
a: Ta có: \(\sqrt{x^2-3x+2}=\sqrt{x-1}\)
\(\Leftrightarrow x^2-3x+2=x-1\)
\(\Leftrightarrow x^2-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)
b: Ta có: \(\sqrt{x^2-4x+4}=\sqrt{4x^2-12x+9}\)
\(\Leftrightarrow\left|x-2\right|=\left|2x-3\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=x-2\\2x-3=-x+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{3}\end{matrix}\right.\)
c: Ta có: \(\sqrt{x^2-5x+6}=\sqrt{x-2}\)
\(\Leftrightarrow x^2-5x+6=x-2\)
\(\Leftrightarrow x^2-6x+8=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
Giaỉ phương trình:
a) \(\sqrt{16\text{x}-48}-6\sqrt{\dfrac{x-3}{4}}+\sqrt{4\text{x}-12}=5\)
b) \(\sqrt{1-10\text{x}+25\text{x}^2}-4=2\)
Cho biểu thức:
A=\(\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}}{1-\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{1-\sqrt{x}}{\sqrt{x}+1}\right)\)
a) Rút gọn A
b) Tính giá trị của biểu thức khi x=\(\frac{2-\sqrt{3}}{2}\)
c) Hãy so sánh A vs \(\frac{1}{2}\)
Các bạn giúp mk vs nhé!!!! ~ Thanks ~
a. ĐK \(x\ge0\)và \(x\ne1\)
A =\(\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}}{1-\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{1-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\frac{\left(\sqrt{x}+1\right)^2+\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\frac{\cdot\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x+2\sqrt{x}+1+x-\sqrt{x}-x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x+2\sqrt{x}+1+\sqrt{x}-x-1+\sqrt{x}}\)
\(=\frac{x+1}{4\sqrt{x}}\)
b. Thay \(x=\frac{2-\sqrt{3}}{2}\Rightarrow A=\frac{\frac{2-\sqrt{3}}{2}+1}{4\sqrt{\frac{2-\sqrt{3}}{2}}}=\frac{4-\sqrt{3}}{4\left(\sqrt{3}-1\right)}=\frac{4-\sqrt{3}}{4-4\sqrt{3}}=-\frac{1+3\sqrt{3}}{8}\)
c . Ta có \(A-\frac{1}{2}=\frac{x+1}{4\sqrt{x}}-\frac{1}{2}=\frac{x-2\sqrt{x}+1}{4\sqrt{x}}=\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}>0\)với \(\forall x>0\)và \(x\ne1\)
Vậy A >1/2
giải các phương trình vô tỉ sau
\(\frac{3}{\sqrt{x}+\sqrt{y}}+\frac{\sqrt{y}}{\sqrt{y}+2}+\frac{\sqrt{y}}{5}+\frac{2}{\sqrt{x}+3}=2\)
\(\sqrt{3x^2-1}+\sqrt{x^2-x}-x\sqrt{x^2+1}=\frac{1}{2\sqrt{2}}\left(7x^2-x+4\right)\)
giúp mình với nhé
Giải phương trình:
a)\(\sqrt{\sqrt{5}-\sqrt{3x}}=\sqrt{8+2\sqrt{15}}\)
b)\(\sqrt{4x-20}-3\sqrt{\dfrac{x-5}{9}}=\sqrt{1-x}\)
c) \(\sqrt{4x+8}+2\sqrt{x+2}-\sqrt{9x+18}=1\)
d) \(\sqrt{x^2-6x+9}+x=11\)
e) \(\sqrt{3x^2-4x+3}=1-2x\)
f) \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)
g) \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)
f) Ta có: \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)
\(\Leftrightarrow4\left|x+1\right|-3\left|x+1\right|=4\)
\(\Leftrightarrow\left|x+1\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
g) Ta có: \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)
\(\Leftrightarrow5\sqrt{x+1}-\sqrt{x+1}=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
Giúp mk bài này với
Giải các phương trình sau:
a, \(\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}}=x\)
b, \(\sqrt{\frac{x^3+1}{x+3}}\sqrt{x+1}=\sqrt{x^2-x+1}+\sqrt{x+3}\)
c, \(\sqrt{x\left(x-1\right)}+\sqrt{x\left(x+2\right)}=2\sqrt{x^2}\)
d, \(\sqrt{x^2-\frac{7}{x^2}}\sqrt{x-\frac{7}{x^2}}=x\)
e, \(\sqrt{\frac{5}{4}-x^2+\sqrt{1-x^2}}+\sqrt{\frac{5}{4}-x^2-\sqrt{1-x^2}}=x+1\)
g, \(\sqrt{7-x^2+x\sqrt{x+5}}=\sqrt{3-2x-x^2}\)
h, \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\frac{x+3}{2}\)
i, \(\sqrt[3]{x+34}-\sqrt[3]{x-3}=1\)
k, \(\sqrt[3]{3x-1}+\sqrt[3]{2x-1}=\sqrt[3]{5x+1}\)
Giải phương trình:
a) \(\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{x^4-1}\)
b) \(3x^2+21x+18+2\sqrt{x^2+7x+7}=2\)
c) \(\sqrt{2-x}+\sqrt{2+x}+\sqrt{4-x^2}=2\)
d) \(\sqrt{9x^2+15x+4}+5\sqrt{4x-7}=5\sqrt{3x+1}+\sqrt{12x^2-5x-28}\)
e) \(\sqrt{x^2-3x+5}+\sqrt{x+4}=\sqrt{x^2-x-1}+\sqrt{2x+1}\)
f) \(\frac{1}{\sqrt{x-1}+\sqrt{x-2}}+\frac{1}{\sqrt{x-2}+\sqrt{x-3}}+...+\frac{1}{\sqrt{x-9}+\sqrt{x-10}}\)
-1; -6
b) ĐK: \(x^2+7x+7\ge0\) (đk xấu quá em ko giải đc;v)
PT \(\Leftrightarrow3x^2+21x+18+2\left(\sqrt{x^2+7x+7}-1\right)=0\)
\(\Leftrightarrow3\left(x+1\right)\left(x+6\right)+2\left(\frac{x^2+7x+6}{\sqrt{x^2+7x+7}+1}\right)=0\)
\(\Leftrightarrow3\left(x+1\right)\left(x+6\right)+\frac{2\left(x+1\right)\left(x+6\right)}{\sqrt{x^2+7x+7}+1}=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)\left[3+\frac{1}{\sqrt{x^2+7x+7}+1}\right]=0\)
Hiển nhiên cái ngoặc vuông > 0 nên vô nghiệm suy ra x = -1 (TM) hoặc x = -6 (TM)
Vậy....
P/s: Cũng may nghiệm đẹp chứ chứ nghiệm xấu thì tiêu rồi:(
chết, đánh nhầm dòng tương đương cuối:
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)\left[3+\frac{2}{\sqrt{x^2+7x+7}+1}\right]=0\)