\(\sqrt{\frac{\gamma+1}{\left(\chi+2\right)^2}}\sqrt{\frac{\left(\chi-2\right)^6}{y^2-2\gamma+1}}\)
Những câu hỏi liên quan
a, A =\(-2\chi^3+3\chi^2+5\) tại \(\chi=\dfrac{1}{2}\);\(\chi^2-1=0\);\(\chi^2=3\chi\)
b, B =\(-3\chi^3\gamma^2+2\chi^2\gamma^3-2\chi\) tại \(\left|\chi\right|=2\);\(\gamma=1\)
a: \(A\left(\dfrac{1}{2}\right)=-2\cdot\dfrac{1}{8}+3\cdot\dfrac{1}{4}+5=\dfrac{11}{2}\)
\(A\left(1\right)=-2+3+5=6\)
\(A\left(-1\right)=2+3+5=10\)
\(A\left(0\right)=-2\cdot0+3\cdot0+5=5\)
\(A\left(-3\right)=-2\cdot\left(-27\right)+3\cdot9+5=86\)
b: Khi x=2 và y=1 thì
\(B=-3\cdot8\cdot1+2\cdot4-2\cdot2=-20\)
Khi x=-2 và y=1 thì
\(B=-3\cdot\left(-8\right)\cdot1+2\cdot4-2\cdot\left(-2\right)=36\)
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trục căn thức và thực hiện phép tính:
Q left(frac{5-2sqrt{5}}{2-sqrt{5}}-2right).left(frac{5+3sqrt{5}}{3+sqrt{5}}-2right)
P frac{3+2sqrt{3}}{sqrt{3}}+frac{2+sqrt{2}}{sqrt{2}+1}-left(sqrt{2}sqrt{3}right)
Mleft(frac{15}{sqrt{6}+1}+frac{4}{sqrt{6}-2}-frac{12}{3-sqrt{6}}right).left(sqrt{6}+11right)
GIẢI CHI TIẾT GIÚP MÌNH NHA!!
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trục căn thức và thực hiện phép tính:
Q= \(\left(\frac{5-2\sqrt{5}}{2-\sqrt{5}}-2\right).\left(\frac{5+3\sqrt{5}}{3+\sqrt{5}}-2\right)\)
P= \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}\sqrt{3}\right)\)
M=\(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right).\left(\sqrt{6}+11\right)\)
GIẢI CHI TIẾT GIÚP MÌNH NHA!!
frac{3+2sqrt{3}}{sqrt{3}}+frac{2+sqrt{2}}{sqrt{2}+1}-left(sqrt{2}+3right)
0.1cdotsqrt{left(-3right)^2}cdotleft[6sqrt{left(frac{1}{3}right)^2}-sqrt{left(sqrt{3}-2right)^2}right]^2
left(frac{3sqrt{2}+sqrt{6}}{sqrt{12}+2}-frac{sqrt{54}}{3}right)cdotfrac{2}{sqrt{6}}
left(frac{3+2sqrt{3}}{sqrt{3}+2}+frac{2+sqrt{2}}{sqrt{2}+1}right)divleft(1divfrac{1}{sqrt{2}+sqrt{3}}right)
sqrt{frac{5+2sqrt{6}}{5-2sqrt{6}}}+sqrt{frac{5-2sqrt{6}}{5+2sqrt{6}}}
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\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)
\(0.1\cdot\sqrt{\left(-3\right)^2}\cdot\left[6\sqrt{\left(\frac{1}{3}\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\right]^2\)
\(\left(\frac{3\sqrt{2}+\sqrt{6}}{\sqrt{12}+2}-\frac{\sqrt{54}}{3}\right)\cdot\frac{2}{\sqrt{6}}\)
\(\left(\frac{3+2\sqrt{3}}{\sqrt{3}+2}+\frac{2+\sqrt{2}}{\sqrt{2}+1}\right)\div\left(1\div\frac{1}{\sqrt{2}+\sqrt{3}}\right)\)
\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)
Dleft ( frac{1}{3sqrt{x}-6} +frac{1}{x-2sqrt{x}}right )left ( frac{1}{6} +frac{1}{2sqrt{x}}right ) Dleft ( frac{1}{3left ( sqrt{x}-2 right )} +frac{1}{sqrt{x}left ( sqrt{x}-2 right )}right ).frac{sqrt{x}+3}{6sqrt{x}} Dfrac{sqrt{x}+3}{3sqrt{x}left ( sqrt{x}-2 right )}.frac{sqrt{x}+3}{6sqrt{x}} Dfrac{left ( sqrt{x}+3 right )^{2}}{18xleft ( sqrt{x}-2 right )} Dfrac{x+6sqrt{x}+9}{18xsqrt{x}-36x}
A/ Đúng
B/ Sai
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\[D=\left ( \frac{1}{3\sqrt{x}-6} +\frac{1}{x-2\sqrt{x}}\right )\left ( \frac{1}{6} +\frac{1}{2\sqrt{x}}\right )\\ D=\left ( \frac{1}{3\left ( \sqrt{x}-2 \right )} +\frac{1}{\sqrt{x}\left ( \sqrt{x}-2 \right )}\right ).\frac{\sqrt{x}+3}{6\sqrt{x}}\\ D=\frac{\sqrt{x}+3}{3\sqrt{x}\left ( \sqrt{x}-2 \right )}.\frac{\sqrt{x}+3}{6\sqrt{x}}\\ D=\frac{\left ( \sqrt{x}+3 \right )^{2}}{18x\left ( \sqrt{x}-2 \right )}\\ D=\frac{x+6\sqrt{x}+9}{18x\sqrt{x}-36x}\]
A/ Đúng
B/ Sai
1) frac{3+2sqrt{3}}{sqrt{3}}+frac{2+sqrt{2}}{sqrt{2}+1}-left(sqrt{2}+3right)2) 0.1sqrt{left(-3right)^2}cdotleft[6sqrt{left(-frac{1}{3}right)^2}-sqrt{left(sqrt{3}-2right)^2}right]^23) left(frac{3+2sqrt{3}}{sqrt{3}+2}+frac{2+sqrt{2}}{sqrt{2}+1}right)divleft(1divfrac{1}{sqrt{2}+sqrt{3}}right)4) left(frac{3sqrt{2}+sqrt{6}}{sqrt{12}+2}-frac{sqrt{54}}{3}right)cdotfrac{2}{sqrt{6}}5) sqrt{frac{5+2sqrt{6}}{5-2sqrt{6}}}+sqrt{frac{5-2sqrt{6}}{5+2sqrt{6}}}
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1) \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)
2) \(0.1\sqrt{\left(-3\right)^2}\cdot\left[6\sqrt{\left(-\frac{1}{3}\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\right]^2\)
3) \(\left(\frac{3+2\sqrt{3}}{\sqrt{3}+2}+\frac{2+\sqrt{2}}{\sqrt{2}+1}\right)\div\left(1\div\frac{1}{\sqrt{2}+\sqrt{3}}\right)\)
4) \(\left(\frac{3\sqrt{2}+\sqrt{6}}{\sqrt{12}+2}-\frac{\sqrt{54}}{3}\right)\cdot\frac{2}{\sqrt{6}}\)
5) \(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)
\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)
\(=\sqrt{\frac{3+2\sqrt{3}\sqrt{2}+2}{3-2\sqrt{3}\sqrt{2}+2}}+\sqrt{\frac{3-2\sqrt{3}\sqrt{2}+2}{3+2\sqrt{3}\sqrt{2}+2}}\)
\(=\sqrt{\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{\left(\sqrt{2}-\sqrt{3}\right)^2}}+\sqrt{\frac{\left(\sqrt{2}-\sqrt{3}\right)^2}{\left(\sqrt{2}+\sqrt{3}\right)^2}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)}+\frac{\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}+\sqrt{3}\right)}\)\
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)+\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)^2+\left(\sqrt{2}-\sqrt{3}\right)^2}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\frac{5+2\sqrt{6}+5-2\sqrt{6}}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=10\)
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\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)
\(=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)
\(=\sqrt{3}+2+\sqrt{2}-\sqrt{2}-3\)
\(=\sqrt{3}-1\)
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14 tháng 1 2022 lúc 9:27
Đố: Cho Delta ABC, biết BCa,ACb,ABc,widehat{A}alpha,widehat{B}beta,widehat{C}gamma chứng minh:a)frac{a}{sinalpha}frac{b}{sinbeta}frac{c}{singamma} b) a^2b^2+c^2-2bccosalphac) frac{a-b}{a+b}frac{tanleft[frac{1}{2}left(alpha-betaright)right]}{tanleft[frac{1}{2}left(alpha+betaright)right]} d) Biết sfrac{a+b+c}{2}. Chứng minh frac{cotfrac{alpha}{2}}{s-a}frac{cotfrac{beta}{2}}{s-b}frac{cotfrac{gamma}{2}}{s-c}
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Đố: Cho \(\Delta ABC\), biết \(BC=a,AC=b,AB=c,\widehat{A}=\alpha,\widehat{B}=\beta,\widehat{C}=\gamma\) chứng minh:
a)\(\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma}\) b) \(a^2=b^2+c^2-2bc\cos\alpha\)
c) \(\frac{a-b}{a+b}=\frac{\tan\left[\frac{1}{2}\left(\alpha-\beta\right)\right]}{\tan\left[\frac{1}{2}\left(\alpha+\beta\right)\right]}\)
d) Biết \(s=\frac{a+b+c}{2}\). Chứng minh \(\frac{\cot\frac{\alpha}{2}}{s-a}=\frac{\cot\frac{\beta}{2}}{s-b}=\frac{\cot\frac{\gamma}{2}}{s-c}\)
Mình rút gọn như thế này đúng không nhỉ?Pleft(2-frac{sqrt{x}-1}{2sqrt{x}-3}right):left(frac{6sqrt{x}+1}{2x-sqrt{x}-3}+frac{sqrt{x}}{sqrt{x}+1}right)Pleft[frac{2left(2sqrt{x}-3right)}{2sqrt{x}-3}-frac{sqrt{x}-1}{2sqrt{x}-3}right]:left[frac{6sqrt{x}+1}{left(sqrt{x}+1right)left(2sqrt{x}-3right)}+frac{sqrt{x}left(2sqrt{x}-3right)}{left(sqrt{x}+1right)left(2sqrt{x}-3right)}right]Pleft(frac{4sqrt{x}-6}{2sqrt{x}-3}-frac{sqrt{x}-1}{2sqrt{x}-3}right):left(frac{6sqrt{x}+1}{left(sqrt{x}+1right)left(2sqrt{x...
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Mình rút gọn như thế này đúng không nhỉ?
\(P=\left(2-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{2x-\sqrt{x}-3}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(P=\left[\frac{2\left(2\sqrt{x}-3\right)}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right]:\left[\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right]\)
\(P=\left(\frac{4\sqrt{x}-6}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\left(\frac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}:\frac{2x+3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\)
\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}.\frac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}{2x+3\sqrt{x}+1}\)
\(P=\left(3\sqrt{x}-5\right).\frac{\left(\sqrt{x}+1\right)}{2x+3\sqrt{x}+1}\)
\(P=\frac{3x+3\sqrt{x}-5\sqrt{x}-5}{2x+3\sqrt{x}+1}\)
\(P=\frac{3x-5\sqrt{x}-5}{2x+1}\)
từ dòng cuối là sai rồi bạn à
Bạn bỏ dòng cuối đi còn lại đúng rồi
Ở tử đặt nhân tử chung căn x chung rồi lại đặt căn x +1 chung
Ở mẫu tách 3 căn x ra 2 căn x +căn x rồi đặt nhân tử 2 căn x ra
rút gọn được \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
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1.\(\left\{{}\begin{matrix}x\left(x-2\right)\left(2x-y\right)=6\\\left(x-3\right)^2+2y=10\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}\frac{1}{\sqrt{x}}+\sqrt{2-\frac{1}{y}}=2\\\frac{1}{\sqrt{y}}+\sqrt{2-\frac{1}{x}}=2\end{matrix}\right.\)
3. \(\left\{{}\begin{matrix}\left(x+1\right)\left(y+1\right)=8\\x\left(x+1\right)+y\left(y+1\right)+xy=17\end{matrix}\right.\)
\(\Rightarrow\left|a\right|\le1\),\(\left|b\right|\le1\),\(\left|c\right|\le1\)
\(\Rightarrow1-a\ge0\)tương tự 1-b,1-c............
\(\Rightarrow\left(1\right)\ge0\)
dấu = khi a=1b=0c=0 và hoán vị
Đang nổi cơn làm biếng mà nhìn thấy hệ còn buồn ngủ hơn:
a/ \(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2-2x\right)\left(2x-y\right)=6\\x^2-2x-2\left(2x-y\right)=1\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x^2-2x=a\\2x-y=b\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}ab=6\\a-2b=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}ab=6\\a=2b+1\end{matrix}\right.\)
\(\Rightarrow b\left(2b+1\right)=6\Leftrightarrow2b^2+b-6=0\)
\(\Leftrightarrow...\)
b/ ĐKXĐ: ...
\(\Leftrightarrow\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{y}}+\sqrt{2-\frac{1}{y}}-\sqrt{2-\frac{1}{x}}=0\)
\(\Leftrightarrow\frac{\sqrt{y}-\sqrt{x}}{\sqrt{xy}}+\frac{\frac{1}{x}-\frac{1}{y}}{\sqrt{2-\frac{1}{y}}+\sqrt{2-\frac{1}{x}}}=0\)
\(\Leftrightarrow\frac{\sqrt{y}-\sqrt{x}}{\sqrt{xy}}+\frac{y-x}{xy\left(\sqrt{2-\frac{1}{y}}+\sqrt{2-\frac{1}{x}}\right)}=0\)
\(\Leftrightarrow\frac{\sqrt{y}-\sqrt{x}}{\sqrt{xy}}+\frac{\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{y}+\sqrt{x}\right)}{xy\left(\sqrt{2-\frac{1}{y}}+\sqrt{2-\frac{1}{x}}\right)}=0\)
\(\Leftrightarrow\left(\sqrt{y}-\sqrt{x}\right)\left(\frac{1}{\sqrt{xy}}+\frac{\sqrt{y}+\sqrt{x}}{xy\left(\sqrt{2-\frac{1}{y}}+\sqrt{2-\frac{1}{x}}\right)}\right)=0\)
\(\Leftrightarrow\sqrt{y}=\sqrt{x}\Rightarrow x=y\)
Thay vào pt đầu:
\(\frac{1}{\sqrt{x}}+\sqrt{2-\frac{1}{x}}=2\)
\(\Leftrightarrow\frac{1}{x}+2-\frac{1}{x}+2\sqrt{\frac{2}{x}-\frac{1}{x^2}}=4\)
\(\Leftrightarrow\sqrt{\frac{2}{x}-\frac{1}{x^2}}=1\)
\(\Leftrightarrow\frac{2}{x}-\frac{1}{x^2}=1\)
\(\Leftrightarrow\left(\frac{1}{x}-1\right)^2=0\)
c/ \(\Leftrightarrow\left\{{}\begin{matrix}xy+x+y=7\\x^2+y^2+x+y+xy=17\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy+x+y=7\\x^2+y^2=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy+x+y=7\\\left(x+y\right)^2-2xy=10\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\) với \(a^2\ge4b\) ta được:
\(\left\{{}\begin{matrix}a+b=7\\a^2-2b=10\end{matrix}\right.\) \(\Rightarrow a^2+2a-24=0\Rightarrow\left[{}\begin{matrix}a=4;b=3\\a=-6;b=13\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;3\right);\left(3;1\right)\)
Xem thêm câu trả lời
tínhleft{left[left(2sqrt{2}right)^2:2,4right].left[5,25:left(sqrt{7}right)^2right]right}:left{left[2frac{1}{7}:frac{left(sqrt{5}right)^2}{7}right]:left[2^3:frac{left(2sqrt{2}right)^2}{sqrt{81}}right]right}tìm x,y,xsqrt{left(x-sqrt{2}right)^2}+sqrt{left(y+sqrt{2}right)^2}+left|x+y+zright|0so sánh A và B:Asqrt{225}-frac{1}{sqrt{5}}-1 Bsqrt{196}-frac{1}{sqrt{6}}ai giải đc câu nào thì giải giúp với
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tính
\(\left\{\left[\left(2\sqrt{2}\right)^2:2,4\right].\left[5,25:\left(\sqrt{7}\right)^2\right]\right\}:\left\{\left[2\frac{1}{7}:\frac{\left(\sqrt{5}\right)^2}{7}\right]:\left[2^3:\frac{\left(2\sqrt{2}\right)^2}{\sqrt{81}}\right]\right\}\)
tìm x,y,x
\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|=0\)
so sánh A và B:
\(A=\sqrt{225}-\frac{1}{\sqrt{5}}-1\) \(B=\sqrt{196}-\frac{1}{\sqrt{6}}\)
ai giải đc câu nào thì giải giúp với