A= \(\sqrt{\frac{x}{4}}\) + \(\sqrt{\frac{4x}{25}}\) -\(\sqrt{\frac{9x}{49}}\)
Những câu hỏi liên quan
giải phương trình vô tỉ
a) \(\frac{3}{4}\sqrt{x}-\sqrt{9x}+5=\frac{1}{4}\sqrt{9x}\)
b) \(\sqrt{3-x}-\sqrt{27-9x}+1,25\sqrt{48-16x}=6\)
c) \(\sqrt{9x^2+12x+4}=4\)
d) \(\frac{1}{3}\sqrt{x-1}+2\sqrt{4x-4}-12\sqrt{\frac{x-1}{25}}=\frac{29}{15}\)
a) \(\frac{3}{4}\sqrt{x}-\sqrt{9x}+5=\frac{1}{4}\sqrt{9x}\)
ĐK : x ≥ 0
⇔ \(\frac{3}{4}\sqrt{x}-\sqrt{3^2x}-\frac{1}{4}\sqrt{3^2x}=-5\)
⇔ \(\frac{3}{4}\sqrt{x}-3\sqrt{x}-\frac{1}{4}\cdot3\sqrt{x}=-5\)
⇔ \(-\frac{9}{4}\sqrt{x}-\frac{3}{4}\sqrt{x}=-5\)
⇔ \(-3\sqrt{x}=-5\)
⇔ \(\sqrt{x}=15\)
⇔ \(x=225\)( tm )
b) \(\sqrt{3-x}-\sqrt{27-9x}+1,25\sqrt{48-16x}=6\)
ĐK : x ≤ 3
⇔ \(\sqrt{3-x}-\sqrt{3^2\left(3-x\right)}+\frac{5}{4}\sqrt{4^2\left(3-x\right)}=6\)
⇔ \(\sqrt{3-x}-3\sqrt{3-x}+\frac{5}{4}\cdot4\sqrt{3-x}=6\)
⇔ \(-2\sqrt{3-x}+5\sqrt{3-x}=6\)
⇔ \(3\sqrt{3-x}=6\)
⇔ \(\sqrt{3-x}=2\)
⇔ \(3-x=4\)
⇔ \(x=-1\)( tm )
c) \(\sqrt{9x^2+12x+4}=4\)
⇔ \(\sqrt{\left(3x+2\right)^2}=4\)
⇔ \(\left|3x+2\right|=4\)
⇔ \(\orbr{\begin{cases}3x+2=4\\3x+2=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-2\end{cases}}\)
d) \(\frac{1}{3}\sqrt{x-1}+2\sqrt{4x-4}-12\sqrt{\frac{x-1}{25}}=\frac{29}{15}\)
ĐK : x ≥ 1
⇔ \(\frac{1}{3}\sqrt{x-1}+2\sqrt{2^2\left(x-1\right)}-12\sqrt{\left(\frac{1}{5}\right)^2\cdot\left(x-1\right)}=\frac{29}{15}\)
⇔ \(\frac{1}{3}\sqrt{x-1}+2\cdot2\sqrt{x-1}-12\cdot\frac{1}{5}\sqrt{x-1}=\frac{29}{15}\)
⇔ \(\frac{1}{3}\sqrt{x-1}+4\sqrt{x-1}-\frac{12}{5}\sqrt{x-1}=\frac{29}{15}\)
⇔ \(\frac{29}{15}\sqrt{x-1}=\frac{29}{15}\)
⇔ \(\sqrt{x-1}=1\)
⇔ \(x-1=1\)
⇔ \(x=2\)( tm )
giúp mik vs
\[ 5\sqrt{\frac{9x - 27}{25}} - 7\sqrt{\frac{4x - 12}{9}} - 7\sqrt{x^2 - 9} + 18\sqrt{\frac{9x^2 - 81}{81}} = 0 \]
=>\(5\cdot\dfrac{3\sqrt{x-3}}{5}-7\cdot\dfrac{2\sqrt{x-3}}{3}-7\cdot\sqrt{x^2-9}+18\cdot\sqrt{\dfrac{9}{81}\left(x^2-9\right)}=0\)
=>\(3\cdot\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}=7\cdot\sqrt{x^2-9}-18\cdot\dfrac{3}{9}\cdot\sqrt{x^2-9}\)
=>\(-\dfrac{5}{3}\sqrt{x-3}=\sqrt{x^2-9}\)
=>\(\sqrt{x-3}\left(\sqrt{x+3}+\dfrac{5}{3}\right)=0\)
=>x-3=0
=>x=3
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Gpt :
1) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)
2) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+s}+\sqrt{x+1}=16\)
3)\(\sqrt{4x+20}+\sqrt{x+5}-\frac{1}{3}\sqrt{9x+45}=4\)
4) \(\frac{1}{3}\sqrt{2x}-\sqrt{8x}+\sqrt{18x}-10=2\)
Bài 1: GIẢI PHƯƠNG TRÌNH
a) \(\sqrt{49-28x-4x^2}-5=0\)
b) \(\frac{1}{2}\sqrt{x-2}-4\sqrt{\frac{4x-8}{9}}-5=0\)
c) \(\sqrt{x^2-4x+4}=7x-1\)
d) \(\frac{1}{5}\sqrt{25x+50}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)
a.
\(DK:49-28x-4x^2\ge0\)
PT\(\Leftrightarrow\sqrt{49-28x-4x^2}=5\)
\(\Leftrightarrow49-28x-4x^2=25\)
\(\Leftrightarrow4x^2+28x-24=0\)
\(\Leftrightarrow x^2+7x-6=0\)
Ta co:
\(\Delta=7^2-4.1.\left(-6\right)=73>0\)
\(\Rightarrow\hept{\begin{cases}x_1=\frac{-7+\sqrt{73}}{2}\left(n\right)\\x_2=\frac{-7-\sqrt{73}}{2}\left(n\right)\end{cases}}\)
Vay nghiem cua PT la \(\hept{\begin{cases}x_1=\frac{-7+\sqrt{73}}{2}\\x_2=\frac{-7-\sqrt{73}}{2}\end{cases}}\)
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Giải phương trình: \(5\sqrt{\frac{9x-27}{25}}-7\sqrt{\frac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\frac{9x^2-81}{91}}=0\)
Tìm x :
h/ sqrt{x+5}-10-4
i/ sqrt{x-5}+2sqrt{4x-20}-frac{1}{3}sqrt{9x-45}12
j/ 3sqrt{2x}+frac{1}{7}sqrt{98x}-sqrt{72x}+40
k/ sqrt{4x^2-20}-frac{1}{3}sqrt{x^2-5}+sqrt{frac{9x^2-45}{16}}-frac{1}{2}sqrt{frac{25x^2-125}{36}}4
l/ sqrt{4x+4}+sqrt{9x+9}-sqrt{x+1}4
m/ sqrt{16left(x+1right)}+sqrt{4x+4}16-sqrt{x+1}+sqrt{9x+9}
Giúp mk với nhé mn
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Tìm x :
h/ \(\sqrt{x+5}-10=-4\)
i/ \(\sqrt{x-5}+2\sqrt{4x-20}-\frac{1}{3}\sqrt{9x-45}=12\)
j/ \(3\sqrt{2x}+\frac{1}{7}\sqrt{98x}-\sqrt{72x}+4=0\)
k/ \(\sqrt{4x^2-20}-\frac{1}{3}\sqrt{x^2-5}+\sqrt{\frac{9x^2-45}{16}}-\frac{1}{2}\sqrt{\frac{25x^2-125}{36}}=4\)
l/ \(\sqrt{4x+4}+\sqrt{9x+9}-\sqrt{x+1}=4\)
m/ \(\sqrt{16\left(x+1\right)}+\sqrt{4x+4}=16-\sqrt{x+1}+\sqrt{9x+9}\)
Giúp mk với nhé mn
h)
ĐKXĐ: $x\geq -5$
PT $\Leftrightarrow \sqrt{x+5}=6$
$\Rightarrow x+5=36\Rightarrow x=31$ (thỏa mãn)
i) ĐKXĐ: $x\geq 5$
PT \(\Leftrightarrow \sqrt{x-5}+4\sqrt{x-5}-\sqrt{x-5}=12\)
\(\Leftrightarrow 4\sqrt{x-5}=12\Leftrightarrow \sqrt{x-5}=3\Rightarrow x-5=9\Rightarrow x=14\) (thỏa mãn)
j)
ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow 3\sqrt{2x}+\sqrt{2x}-6\sqrt{2x}+4=0$
$\Leftrightarrow -2\sqrt{2x}+4=0$
$\Leftrightarrow \sqrt{2x}=2$
$\Rightarrow x=2$ (thỏa mãn)
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k) ĐK: $x^2\geq 5$
PT $\Leftrightarrow 2\sqrt{x^2-5}-\frac{1}{3}\sqrt{x^2-5}+\frac{3}{4}\sqrt{x^2-5}-\frac{5}{12}\sqrt{x^2-5}=4$
$\Leftrightarrow 2\sqrt{x^2-5}=4$
$\Leftrightarrow \sqrt{x^2-5}=2$
$\Rightarrow x^2-5=4$
$\Leftrightarrow x^2=9\Rightarrow x=\pm 3$ (đều thỏa mãn)
l) ĐKXĐ: $x\geq -1$
PT $\Leftrightarrow 2\sqrt{x+1}+3\sqrt{x+1}-\sqrt{x+1}=4$
$\Leftrightarrow 4\sqrt{x+1}=4$
$\Leftrightarrow \sqrt{x+1}=1$
$\Rightarrow x+1=1$
$\Rightarrow x=0$
m)
ĐKXĐ: $x\geq -1$
PT $\Leftrightarrow 4\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}+3\sqrt{x+1}$
$\Leftrightarrow 6\sqrt{x+1}=16+2\sqrt{x+1}$
$\Leftrightarrow 4\sqrt{x+1}=16$
$\Leftrightarrow \sqrt{x+1}=4$
$\Rightarrow x=15$ (thỏa mãn)
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Tìm x, biết:
a, \(\sqrt{4x+20}-3\sqrt{5+x}+\frac{4}{3}\sqrt{9x+45}=6\)
b, \(\sqrt{25x-25}-\frac{15}{2}\sqrt{\frac{x-1}{9}}=6+\sqrt{x-1}\)
a. \(\Rightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\Rightarrow\sqrt{x+5}\left(2-3+4\right)=6\Rightarrow\sqrt{x+5}=2\Rightarrow x+5=4\Rightarrow x=-1\)
b.\(\Rightarrow5\sqrt{x-1}-\frac{5}{2}\sqrt{x-1}-\sqrt{x-1}=6\Rightarrow\sqrt{x-1}\left(5-\frac{5}{2}-1\right)=6\Rightarrow\sqrt{x-1}=4\Rightarrow x-1=16\Rightarrow x=17\)
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Mình làm ý a thôi nha còn ý b tương tự
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Giải phương trình
a \(\sqrt{x^2-4}-3\sqrt{x-2}=0\)
b \(x-6\sqrt{x}+9=0\)
c \(\sqrt{9x-27}+\sqrt{x-3}-\frac{1}{2}\sqrt{4x-12}=7\)
d \(3\sqrt{8x+4}-\frac{1}{3}\sqrt{18x+9}-\frac{1}{2}\sqrt{50x+25}+\sqrt[]{\frac{2x+1}{4}}=6\)
Lời giải:
a) ĐK: $x\geq 2$
PT $\Leftrightarrow \sqrt{(x-2)(x+2)}-3\sqrt{x-2}=0$
$\Leftrightarrow \sqrt{x-2}(\sqrt{x+2}-3)=0$
\(\Rightarrow \left[\begin{matrix} \sqrt{x-2}=0\\ \sqrt{x+2}-3=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=2\\ x=7\end{matrix}\right.\) (thỏa mãn)
Vậy..........
b) ĐK: $x\geq 0$
PT $\Leftrightarrow (\sqrt{x}-3)^2=0$
$\Leftrightarrow \sqrt{x}-3=0$
$\Leftrightarrow x=9$ (thỏa mãn)
c) ĐK: $x\geq 3$
PT $\Leftrightarrow \sqrt{9(x-3)}+\sqrt{x-3}-\frac{1}{2}\sqrt{4(x-3)}=7$
$\Leftrightarrow 3\sqrt{x-3}+\sqrt{x-3}-\sqrt{x-3}=7$
$\Leftrightarrow 3\sqrt{x-3}=7$
$\Leftrightarrow x-3=(\frac{7}{3})^2$
$\Rightarrow x=\frac{76}{9}$
d)
ĐK: $x\geq \frac{-1}{2}$
PT $\Leftrightarrow 3\sqrt{4(2x+1)}-\frac{1}{3}\sqrt{9(2x+1)}-\frac{1}{2}\sqrt{25(2x+1)}+\sqrt{\frac{1}{4}(2x+1)}=6$
$\Leftrightarrow 6\sqrt{2x+1}-\sqrt{2x+1}-\frac{5}{2}\sqrt{2x+1}+\frac{1}{2}\sqrt{2x+1}=6$
$\Leftrightarrow 3\sqrt{2x+1}=6$
$\Leftrightarrow \sqrt{2x+1}=2$
$\Rightarrow x=\frac{3}{2}$ (thỏa mãn)
BÀI 1: RÚT GỌN
1)frac{1}{sqrt{3}+1}+frac{1}{sqrt{3}-1}
2)sqrt{7+2sqrt{10}}+2sqrt{frac{1}{5}}-frac{1}{sqrt{5}-2}
3)frac{3}{sqrt{3}-1}+sqrt{frac{4}{3}}-sqrt{8+2sqrt{5}}
4)3sqrt{frac{16x}{81}}+frac{5}{4}sqrt{frac{4x}{25}}-frac{2}{x}sqrt{frac{9a^3}{4}}
5)frac{1}{3}sqrt{3a}-frac{2}{3}sqrt{frac{27a}{4}}+frac{5}{a}sqrt{frac{12a^3}{5}}
BÀI 2: GIẢI PHƯƠNG TRÌNH
1)sqrt{5x-1}sqrt{2}-1
2)sqrt{1-2x}sqrt{3}-1
3)4sqrt{x}-2sqrt{9x}+sqrt{16x}20
4)frac{3}{5}sqrt{frac{25x-75}{16}}-frac{1}{14}sqrt{49x-147}20...
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BÀI 1: RÚT GỌN
1)\(\frac{1}{\sqrt{3}+1}+\frac{1}{\sqrt{3}-1}\)
2)\(\sqrt{7+2\sqrt{10}}+2\sqrt{\frac{1}{5}}-\frac{1}{\sqrt{5}-2}\)
3)\(\frac{3}{\sqrt{3}-1}+\sqrt{\frac{4}{3}}-\sqrt{8+2\sqrt{5}}\)
4)\(3\sqrt{\frac{16x}{81}}+\frac{5}{4}\sqrt{\frac{4x}{25}}-\frac{2}{x}\sqrt{\frac{9a^3}{4}}\)
5)\(\frac{1}{3}\sqrt{3a}-\frac{2}{3}\sqrt{\frac{27a}{4}}+\frac{5}{a}\sqrt{\frac{12a^3}{5}}\)
BÀI 2: GIẢI PHƯƠNG TRÌNH
\(1)\sqrt{5x-1}=\sqrt{2}-1\\ 2)\sqrt{1-2x}=\sqrt{3}-1\\ 3)4\sqrt{x}-2\sqrt{9x}+\sqrt{16x}=20\\ 4)\frac{3}{5}\sqrt{\frac{25x-75}{16}}-\frac{1}{14}\sqrt{49x-147}=20\\ 5)\frac{1}{2}\sqrt{x-2}-4\sqrt{\frac{4x-8}{9}}+\sqrt{9x-18}-5=0\)
BÀI 3: CHO BIỂU THỨC
Q=\(\frac{2}{2+\sqrt{x}}+\frac{1}{2-\sqrt{x}}+\frac{2\sqrt{x}}{x-4}\) ĐKXĐ x ≥ 0, x ≠ 4
a) Rút gọn biểu thức Q
b) Tính Q thì x = 81
c) Tìm x để Q = \(\frac{6}{5}\)
d) Tìm x để nguyên đó Q nguyên