tìm x biết
a)\(\frac{3\sqrt{x}-5}{2}-\frac{2\sqrt{x}-7}{3}+1=\sqrt{x}\)
b)\(\sqrt{9x^2+45}-\frac{1}{12}\sqrt{16x^2+80}+3\sqrt{\frac{x^2+5}{16}}-\frac{1}{4}\sqrt{\frac{25x^2+125}{9}}=9\)
Những câu hỏi liên quan
\(\frac{1}{4}\sqrt{\frac{25x^2+125}{9}}\)Tìm x biết
a) \(\frac{3\sqrt{x}-5}{2}\)- \(\frac{2\sqrt{x}-7}{3}\)+1=20
b) \(\sqrt{9x^2+45}\) - \(\frac{1}{12}\sqrt{16x^2+80}\) +\(3\sqrt{\frac{x^2+5}{16}}\)
-\(\frac{1}{4}\sqrt{\frac{25x^2+125}{9}}\)=9
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
Đọc tiếp
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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Asqrt{80}+sqrt{45}+sqrt{5}Bfrac{5}{sqrt{10}}+3,5.sqrt{40}Cfrac{1}{sqrt{3}-2}+frac{sqrt{300}}{10}-sqrt{12}D4sqrt{x}+2sqrt{x^2}-sqrt{16x}( x hoặc 0 )Esqrt{25x+25}-sqrt{9x+9}+sqrt{4x+x}vớixge-1Ffrac{a-2sqrt{a}}{sqrt{a}-2}vớiage0,ne4Gfrac{2}{sqrt{3}+sqrt{5}}-frac{2}{sqrt{5}-sqrt{7}}
Đọc tiếp
\(A=\sqrt{80}+\sqrt{45}+\sqrt{5}\)
\(B=\frac{5}{\sqrt{10}}+3,5.\sqrt{40}\)
\(C=\frac{1}{\sqrt{3}-2}+\frac{\sqrt{300}}{10}-\sqrt{12}\)
\(D=4\sqrt{x}+2\sqrt{x^2}-\sqrt{16x}\)( x > hoặc = 0 )
\(E=\sqrt{25x+25}-\sqrt{9x+9}+\sqrt{4x+x}vớix\ge-1\)
\(F=\frac{a-2\sqrt{a}}{\sqrt{a}-2}vớia\ge0,\ne4\)
\(G=\frac{2}{\sqrt{3}+\sqrt{5}}-\frac{2}{\sqrt{5}-\sqrt{7}}\)
Đề bài là Rút gọn biểu thức nha . Mình quên ghi ^^
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\(A=\sqrt{80}+\sqrt{45}+\sqrt{5}=\sqrt{16.5}+\sqrt{9.5}+\sqrt{5}\)
\(=4\sqrt{5}+3\sqrt{5}+\sqrt{5}=8\sqrt{5}\)
\(B=\frac{5}{\sqrt{10}}+3,5\sqrt{40}=\sqrt{\frac{25}{10}}+3,5\sqrt{16.2,5}\)
\(=\sqrt{2,5}+3,5.4\sqrt{2,5}=15\sqrt{2,5}\)
\(C=\frac{1}{\sqrt{3}-2}+\frac{\sqrt{300}}{10}-\sqrt{12}\)
\(=\frac{\sqrt{3}+2}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}+\frac{\sqrt{100.3}}{10}-\sqrt{4.3}\)
\(=-\sqrt{3}-2+\sqrt{3}-2\sqrt{3}=-2\sqrt{3}-2\)
\(D=4\sqrt{x}+2\sqrt{x^2}-\sqrt{16x}=4\sqrt{x}+2x-4\sqrt{x}=2x\) ( do \(x\ge0\))
\(F=\frac{a-2\sqrt{a}}{\sqrt{a}-2}=\frac{\sqrt{a}.\left(\sqrt{a}-2\right)}{\sqrt{a}-2}=\sqrt{a}\)
mk chỉnh đề
\(E=\sqrt{25x+25}-\sqrt{9x+9}+\sqrt{4x+4}\)
\(=\sqrt{25\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}\)
\(=5\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}=4\sqrt{x+1}\)
\(G=\frac{2}{\sqrt{3}+\sqrt{5}}-\frac{2}{\sqrt{5}-\sqrt{7}}=\frac{2\left(\sqrt{3}-\sqrt{5}\right)}{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{3}-\sqrt{5}\right)}-\frac{2\left(\sqrt{5}+\sqrt{7}\right)}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{5}-\sqrt{7}\right)}\)
\(=\sqrt{3}-\sqrt{5}-\sqrt{5}-\sqrt{7}=\sqrt{3}-\sqrt{7}\)
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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...
Đọc tiếp
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
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) \(\frac{1}{3}\sqrt{2x}-\sqrt{8x}+\sqrt{18x}-10=2\)
a)sqrt{x^2+2x+10}+x^2+2x+80
b)15x-2x^2-5sqrt{2x^2-15x+11}
c)sqrt{9x^2+45}+sqrt{16x^2+80}+3sqrt{frac{x^2+5}{16}}-frac{1}{4}sqrt{frac{25x^2+15}{9}}9
d)3x^2+21x+18+2sqrt{x^2+7x+7}2
e)sqrt{x^2+3x+2}-2sqrt{2x^2+6x+2}-sqrt{2}
f)sqrt{x-1}+sqrt{x+3}-sqrt{x^2+2x-3}-10
Đọc tiếp
a)\(\sqrt{x^2+2x+10}+x^2+2x+8=0\)
b)\(15x-2x^2-5=\sqrt{2x^2-15x+11}\)
c)\(\sqrt{9x^2+45}+\sqrt{16x^2+80}+3\sqrt{\frac{x^2+5}{16}}-\frac{1}{4}\sqrt{\frac{25x^2+15}{9}}=9\)
d)\(3x^2+21x+18+2\sqrt{x^2+7x+7}=2\)
e)\(\sqrt{x^2+3x+2}-2\sqrt{2x^2+6x+2}=-\sqrt{2}\)
f)\(\sqrt{x-1}+\sqrt{x+3}-\sqrt{x^2+2x-3}-1=0\)
a) + \(VT=\sqrt{x^2+2x+10}+x^2+2x+1+7\)
\(=\sqrt{x^2+2x+1}+\left(x+1\right)^2+7>0\forall x\)
=> ptvn
d) ĐK : \(x^2+7x+7\ge0\)
Đặt \(t=\sqrt{x^2+7x+7}\ge0\) \(\Rightarrow t^2=x^2+7x+7\)
\(pt\Leftrightarrow3\left(x^2+7x+7\right)-3+2\sqrt{x^2+7x+7}-2=0\)
\(\Leftrightarrow3t^2+2t-5=0\Leftrightarrow\left(3t+5\right)\left(t-1\right)=0\)
\(\Leftrightarrow t=1\) ( do \(3t+5>0\forall t\ge0\) )
\(\Leftrightarrow x^2+7x+1=0\Leftrightarrow x^2+7x+6=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\) ( TM )
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f) ĐK : \(x\ge1\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-1}\ge0\\b=\sqrt{x+3}\ge0\end{matrix}\right.\) thì pt trở thành :
\(a+b-ab-1=0\)
\(\Leftrightarrow\left(a-1\right)-b\left(a-1\right)=0\)
\(\Leftrightarrow\left(1-b\right)\left(a-1\right)=0\Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x+3}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\x=-2\left(KTM\right)\end{matrix}\right.\)
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Tìm x :
a, sqrt{x^2-2x}sqrt{2-3x}
b, sqrt{x-3}-2sqrt{x^2-9}0
c, sqrt{4x-20}+sqrt{x-5}-frac{1}{3}sqrt{9x-45}4
d, frac{1}{2}sqrt{x-1}-frac{3}{2}sqrt{9x-9}+24sqrt{frac{x-1}{64}}-17
e, sqrt{9x^2+18}+2sqrt{x^2+2}-sqrt{25x^2+50}+30
f, sqrt{x^2-4}-x+20
Đọc tiếp
Tìm x :
a, \(\sqrt{x^2-2x}=\sqrt{2-3x}\)
b, \(\sqrt{x-3}-2\sqrt{x^2-9}=0\)
c, \(\sqrt{4x-20}+\sqrt{x-5}-\frac{1}{3}\sqrt{9x-45}=4\)
d, \(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)
e, \(\sqrt{9x^2+18}+2\sqrt{x^2+2}-\sqrt{25x^2+50}+3=0\)
f, \(\sqrt{x^2-4}-x+2=0\)
a/\(\sqrt{x^2-2x}=\sqrt{2-3x}\left(đk:x\le0\right)
\)
\(\Leftrightarrow x^2-2x=2-3x\)
\(\Leftrightarrow x^2+x-2=0\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(KTM\right)\\x=-2\left(TM\right)\end{matrix}\right.\)
Vậy x=-2 là nghiệm của PT
b/\(\sqrt{x-3}-2\sqrt{x^2-9}=0\left(đk:x\ge3\right)\)
\(\Leftrightarrow\sqrt{x-3}\left(1-2\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\1=2\sqrt{x+3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(TM\right)\\4x+12=1\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=3\\x=-\frac{11}{4}\left(KTM\right)\end{matrix}\right.\)
Vậy x=3
Bài 1: Tìm x, biết
a)\(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)
b) \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
c)\(\sqrt{16x-16}-\sqrt{9x-9}+\sqrt{4x-4}+\sqrt{x-1}=8\)
d) \(\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}=2\)
a) Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)
\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)
\(\Leftrightarrow4\sqrt{x-3}=20\)
\(\Leftrightarrow x-3=25\)
hay x=28
b) Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)
\(\Leftrightarrow2\sqrt{x+2}=6\)
\(\Leftrightarrow x+2=9\)
hay x=7
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Giải phương trình
a,\(\sqrt{x^2-2x+1}=2x\)
b,\(\sqrt{25x-125}-3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=6\)
\(a,ĐKXĐ:x\ge0\\ \Leftrightarrow\sqrt{\left(x-1\right)^2}=2x\\ \Leftrightarrow\left|x-1\right|=2x\\ \Rightarrow\left[{}\begin{matrix}x-1=2x\\1-x=2x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(kot/mĐKXĐ\right)\\x=\frac{1}{3}\left(t/m\right)\end{matrix}\right.\\ Vậy.....\)
\(b,ĐKXĐ:x\ge5\\ \Leftrightarrow\sqrt{25\left(x-5\right)}-3\cdot\frac{1}{3}\cdot\sqrt{x-5}-\frac{1}{3}\cdot3\cdot\sqrt{x-5}\Leftrightarrow5\sqrt{x-5}-\sqrt{x-5}-\sqrt{x-5}=6\\ \Leftrightarrow\left(5-1-1\right)\sqrt{x-5}=6\\ \Leftrightarrow\sqrt{x-5}=2\\ \Rightarrow x-5=4\\ \Leftrightarrow x=9\left(thỏamãnĐKXĐ\right)\\ Vậy...\)
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