a) \(^{â^3-2\sqrt{2}}\)
b)\(\sqrt{21}+\sqrt{3}+\sqrt{7}+1\)
c) \(x+2\sqrt{x}-3\)
Những câu hỏi liên quan
Rút gọn các biểu thức sau
a) $M=\sqrt{\dfrac{3 a}{7}}-2 \sqrt{\dfrac{7 a}{3}}+\sqrt{21 a};$
b) $N=\sqrt{\dfrac{8 x}{3}}-\sqrt{\dfrac{27 x}{2}}+\sqrt{6 x};$
c) $P=2 \sqrt{\dfrac{8 y}{5}}+\sqrt{\dfrac{45 y}{2}}-\sqrt{10 y}$.
a)\(\sqrt{\frac{3a}{7}}-2\sqrt{\frac{7a}{3}}+\sqrt{21a}\) =\(\sqrt{\frac{3}{7}.\frac{1}{21}.21a}\) - \(2\sqrt{\frac{7}{3}.\frac{1}{21}.21a}\)+ \(\sqrt{21}\)
=\(\sqrt{\frac{1}{49}.21a}\) - \(2\sqrt{\frac{1}{9}.21a}\)+\(\sqrt{21}\)
=\(\sqrt{\frac{1}{49}}.\sqrt{21a}\) - \(2.\sqrt{\frac{1}{9}}.\sqrt{21a}\)+ \(\sqrt{21a}\)
=\(\frac{1}{7}\sqrt{21a}\) - \(\frac{2}{3}\sqrt{21a}\) + \(\sqrt{21a}\)
=\(\frac{-10}{21}\sqrt{21a}\)
b)
N=\(\sqrt{\frac{8x}{3}}\) - \(\sqrt{\frac{27x}{2}}\) + \(\sqrt{6x}\)
=\(\sqrt{\frac{8}{3}.\frac{1}{6}.6x}\) - \(\sqrt{\frac{27}{2}.\frac{1}{6}.6x}\)+ \(\sqrt{6x}\)
=\(\frac{2}{3}\sqrt{6x}-\frac{3}{2}.\sqrt{6x}+\sqrt{6x}\)
=\(\frac{1}{6}\sqrt{6x}\)
em lớp 8 nene làm theo cách hiểu thôi ạ
c)P=\(2\sqrt{\frac{8y}{5}}\) + \(\sqrt{\frac{45y}{2}}\) - \(\sqrt{10y}\)
=\(2\sqrt{\frac{8}{5}.\frac{1}{10}.10y}\) + \(\sqrt{\frac{45}{2}.\frac{1}{10}.10y}\) - \(\sqrt{10y}\)
=\(2\sqrt{\frac{4}{25}.10y}\) + \(\sqrt{\frac{9}{4}.10y}\) - \(\sqrt{10y}\)
=\(2\).\(\sqrt{\frac{4}{25}}\) \(.\sqrt{10y}\) + \(\sqrt{\frac{9}{4}}.\sqrt{10y}\) - \(\sqrt{10y}\)
=\(\frac{4}{5}\sqrt{10y}\) + \(\frac{3}{2}\sqrt{10y}\) - \(\sqrt{10y}\)
=\(\frac{13}{10}\sqrt{10y}\)
Adfrac{sqrt{x}+2}{sqrt{x}-2}-dfrac{3}{sqrt{x}+2}+dfrac{12}{x-4}Bdfrac{sqrt{x}}{sqrt{x}-3}-dfrac{sqrt{x}-21}{9-x}dfrac{1}{sqrt{x}+3}Cdfrac{sqrt{x}}{sqrt{x}+3}+dfrac{2sqrt{x}}{sqrt{x}-3}-dfrac{3x+9}{x-9}Ddfrac{1}{sqrt{x}+3}-dfrac{sqrt{x}}{3-sqrt{x}}+dfrac{2sqrt{x}+12}{x-9}Ndfrac{sqrt{x}}{sqrt{x}-1}+dfrac{2sqrt{x}-1}{sqrt{x}+1}-dfrac{6}{x-1}Mdfrac{3}{sqrt{x}-3}+dfrac{2}{sqrt{x}+3}+dfrac{x-5sqrt{x}-3}{x-9}
Đọc tiếp
\(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{3}{\sqrt{x}+2}+\dfrac{12}{x-4}\)
\(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{\sqrt{x}-21}{9-x}\dfrac{1}{\sqrt{x}+3}\)
\(C=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
\(D=\dfrac{1}{\sqrt{x}+3}-\dfrac{\sqrt{x}}{3-\sqrt{x}}+\dfrac{2\sqrt{x}+12}{x-9}\)
\(N=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{6}{x-1}\)
\(M=\dfrac{3}{\sqrt{x}-3}+\dfrac{2}{\sqrt{x}+3}+\dfrac{x-5\sqrt{x}-3}{x-9}\)
a: Ta có: \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{3}{\sqrt{x}+2}+\dfrac{12}{x-4}\)
\(=\dfrac{x+4\sqrt{x}+4-3\sqrt{x}+6+12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+\sqrt{x}+22}{x-4}\)
d: Ta có: \(D=\dfrac{1}{\sqrt{x}+3}-\dfrac{\sqrt{x}}{3-\sqrt{x}}+\dfrac{2\sqrt{x}-12}{x-9}\)
\(=\dfrac{\sqrt{x}-3+x+3\sqrt{x}+2\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x+6\sqrt{x}-15}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
Đúng 0
Bình luận (0)
A=\(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{3}{\sqrt{x}+2}+\dfrac{12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}A=\dfrac{\sqrt{x}+2.\left(\sqrt{x}+2\right)-3.\left(\sqrt{x}-2\right)+12}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+2\right)}A=\dfrac{\sqrt{x}+2\sqrt{x}+4-3\sqrt{x}+6+12}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+2\right)}A=\dfrac{22}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+2\right)}\)
Đúng 0
Bình luận (0)
giải pt :
a,\(2x^2-11x+21=3\sqrt[3]{4x-4}\)
b,\(\dfrac{\sqrt{x-3}}{\sqrt{2x-1}-1}=\dfrac{1}{\sqrt{x+3}-\sqrt{x-3}}\)
c,\(\left(\sqrt{x^2+x+1}+\sqrt{4x^2+x+1}\right)\left(\sqrt{5x^2+1}-\sqrt{2x^2+1}\right)=3x^2\)
Giải pt:
a) xsqrt{1-dfrac{1}{x}}+sqrt{x-dfrac{1}{x}}
b) sqrt{x^2+x}+sqrt{x-x^2}x+1
c) sqrt{x^2-x}+sqrt{x^2+2x}2sqrt{x^2}
d)sqrt{dfrac{x^3+1}{x+3}}+sqrt{x+1}sqrt{x^2-x+1}+sqrt{x+3}
e) sqrt{sqrt{3}-x}xsqrt{sqrt{3}+x}
f) 4xsqrt{x+7}+3xsqrt{7x-3}6x^2+2sqrt{7x^2+46x-21}
Đọc tiếp
Giải pt:
a) x=\(\sqrt{1-\dfrac{1}{x}}+\sqrt{x-\dfrac{1}{x}}\)
b) \(\sqrt{x^2+x}+\sqrt{x-x^2}=x+1\)
c) \(\sqrt{x^2-x}+\sqrt{x^2+2x}=2\sqrt{x^2}\)
d)\(\sqrt{\dfrac{x^3+1}{x+3}}+\sqrt{x+1}=\sqrt{x^2-x+1}+\sqrt{x+3}\)
e) \(\sqrt{\sqrt{3}-x}=x\sqrt{\sqrt{3}+x}\)
f) \(4x\sqrt{x+7}+3x\sqrt{7x-3}=6x^2+2\sqrt{7x^2+46x-21}\)
a) ĐKXĐ: \(\left[{}\begin{matrix}x\ge1\\0>x\ge-1\end{matrix}\right.\). Để pt có nghiệm => x>0=> \(x\ge1\) pt<=> \(x-\sqrt{1-\dfrac{1}{x}}=\sqrt{x-\dfrac{1}{x}}.Bìnhphương2vetaco\left(x-\sqrt{1-\dfrac{1}{x}}\right)^2=x-\dfrac{1}{x}\)\(\Leftrightarrow x^2+1-\dfrac{1}{x}-2x\sqrt{1-\dfrac{1}{x}}=x-\dfrac{1}{x}\Leftrightarrow x^2-x+1=2\sqrt{x^2-x}\Leftrightarrow\left(\sqrt{x^2-x}-1\right)^2=0\Leftrightarrow x^2-x=1\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{5}{4}\)
Đúng 0
Bình luận (0)
b) ĐKXĐ\(0\le x\le1\) pt \(\Leftrightarrow\left(\sqrt{x^2+x}+\sqrt{x-x^2}\right)^2=\left(x+1\right)^2\Leftrightarrow2x+2x.\sqrt{1-x^2}=x^2+2x+1\Leftrightarrow x^2-2x\sqrt{1-x^2}+1-x^2+x^2=0\Leftrightarrow\left(x-\sqrt{1-x^2}\right)^2+x^2=0\)
Đúng 0
Bình luận (0)
c)ĐKXĐ:x=0 hoặc \(x\ge1;x\le-2\)
Nếu x=0=> VT=VP=0=> x=0 là 1 no
Nếu \(x\ge1.\)pt<=>\(\sqrt{x-1}+\sqrt{x+2}=2\sqrt{x}\Leftrightarrow x-1+x+2+2\sqrt{x^2+x-2}=4x\Leftrightarrow2x-1=2\sqrt{x^2+x-2}\Leftrightarrow4x^2-4x+1=4\left(x^2+x-2\right)\left(Dox\ge1\right)\)\(\Leftrightarrow8x=9\)=>....
Nếu \(x\le-2.\)Chia cả 2 vế của pt cho \(\sqrt{-x}\).Giải tương tự x>=1
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
2 tháng 6 2021 lúc 8:27
a.\(\sqrt{28a^4}\)
b. A=\(\left(\dfrac{\sqrt{21}-\sqrt{7}}{\sqrt{3-1}}+\dfrac{\sqrt{10}-\sqrt{5}}{\sqrt{2}-1}\right)\)\(\div\)\(\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
c.\(\left\{{}\begin{matrix}\dfrac{3}{2x}-y=6\\\dfrac{1}{x}+2y=-4\end{matrix}\right.\)
`a)sqrt{28a^4}`
`=sqrt{7.4.a^4}`
`=2sqrt7a^2`
`b)A=((sqrt{21}-sqrt7)/(sqrt3-1)+(sqrt{10}-sqrt5)/(sqrt2-1)):1/(sqrt7-sqrt5)`
`=((sqrt7(sqrt3-1))/(sqrt3-1)+(sqrt5(sqrt2-1))/(sqrt2-1)).(sqrt7-sqrt5)`
`=(sqrt7+sqrt5)(sqrt7-sqrt5)`
`=7-5=2`
`c)` $\begin{cases}\dfrac{3}{2x}-y=6\\\dfrac{1}{x}+2y=-4\end{cases}$
`<=>` $\begin{cases}\dfrac{3}{x}-2y=12\\\dfrac{1}{x}+2y=-4\end{cases}$
`<=>` $\begin{cases}\dfrac{4}{x}=8\\2y+\dfrac{1}{x}=-4\end{cases}$
`<=>` $\begin{cases}x=\dfrac12\\2y=-4-2=-6\end{cases}$
`<=>` $\begin{cases}x=\dfrac12\\y=-3\end{cases}$
Vậy HPT có nghiệm `(x,y)=(1/2,-3)`.
Đúng 1
Bình luận (1)
Phương pháp 7. Nhẩm nghiệm và biến đổi về phương trình tích, có sử dụng phép nhân
liên hợp
a \(\sqrt{x-2}+\sqrt{x+1}+\sqrt{2x+3}=6\)
b \(x^2+5\sqrt{x-3}=21\)
c \(x^2+4x+\sqrt{4x+5}+\sqrt{x+3}-10=0\)
b. Tự đặt đk
\(x^{^2}+5\sqrt{x-3}=21\\\Leftrightarrow x^{^2}-9+5\sqrt{x-3}=12 \)
Đặt \(a=\sqrt{x-3}\) \(\left(a\ge0\right)\) Phương trình trở thành:
\(a^{^2}\left(a^{^2}+6\right)+5a=12\\ \Leftrightarrow a^{^4}+6a^{^2}+5a-12=0\\ \Leftrightarrow a^{^4}-a^{^3}+a^{^3}-a^{^2}+7a^{^2}-7a+12a-12=0\\ \Leftrightarrow\left(a-1\right)\left(a^{^3}+a^{^2}+7a+12\right)=0\\ \Leftrightarrow a=1\left(tmdk\right)\)
Ta có: vì \(a\ge0\) nên \(a^{^3}+a^{^2}+7a+12\ne0\)
Với a = 1 ta có x=4 (tmdk)
Đúng 1
Bình luận (1)
Rút gọn biểu thức 1) dfrac{sqrt{14}-sqrt{21}}{sqrt{7}} .2) dfrac{sqrt{a^2+5a+6}}{sqrt{a+3}}3) sqrt{3left(x^2-10x+25right)}.sqrt{27} với x 54)dfrac{y}{x}sqrt{dfrac{x^2}{y^4}} với x 0; y 05) dfrac{1}{x-y}.sqrt{x^6left(x-yright)^4} với x ne y
Đọc tiếp
Rút gọn biểu thức 1) \(\dfrac{\sqrt{14}-\sqrt{21}}{\sqrt{7}}\) .
2) \(\dfrac{\sqrt{a^2+5a+6}}{\sqrt{a+3}}\)
3) \(\sqrt{3\left(x^2-10x+25\right)}.\sqrt{27}\) với x < 5
4)
\(\dfrac{y}{x}\sqrt{\dfrac{x^2}{y^4}}\) với x > 0; y < 0
5) \(\dfrac{1}{x-y}.\sqrt{x^6\left(x-y\right)^4}\) với x \(\ne\) y
5: \(=\dfrac{1}{x-y}\cdot x^3\cdot\left(x-y\right)^2=x^3\left(x-y\right)\)
Đúng 1
Bình luận (0)
bài 1) rút gọn
1) 5√frac{1}{5} 2)frac{12}{5}√frac{5}{4} 3)frac{30}{5sqrt{6}} 4) frac{20}{2sqrt{5}} 5)frac{2-sqrt{2}}{sqrt{2}} 6) frac{11+sqrt{11}}{1+sqrt{ }11} 7) frac{sqrt{21-sqrt{7}}}{1-sqrt{3}} 8)frac{sqrt{2+sqrt{3}}}{2+sqrt{6}} 9)frac{sqrt{10-sqrt{2}}}{sqrt{5-}1} 10)frac{2sqrt{3}-3sqrt{2}}{sqrt{3}-sqrt[]{2}}
bài 2) với các biểu thức đã cho là có nghĩa và rút gọn
1)frac{x-sqrt{x}}{sqrt{x}-1} 2)frac{xsqrt{x}-2x}{2-sqrt{x}} 3) frac{xsqrt{y}-...
Đọc tiếp
bài 1) rút gọn
1) 5√\(\frac{1}{5}\) 2)\(\frac{12}{5}\)√\(\frac{5}{4}\) 3)\(\frac{30}{5\sqrt{6}}\) 4) \(\frac{20}{2\sqrt{5}}\) 5)\(\frac{2-\sqrt{2}}{\sqrt{2}}\) 6) \(\frac{11+\sqrt{11}}{1+\sqrt{ }11}\) 7) \(\frac{\sqrt{21-\sqrt{7}}}{1-\sqrt{3}}\) 8)\(\frac{\sqrt{2+\sqrt{3}}}{2+\sqrt{6}}\) 9)\(\frac{\sqrt{10-\sqrt{2}}}{\sqrt{5-}1}\) 10)\(\frac{2\sqrt{3}-3\sqrt{2}}{\sqrt{3}-\sqrt[]{2}}\)
bài 2) với các biểu thức đã cho là có nghĩa và rút gọn
1)\(\frac{x-\sqrt{x}}{\sqrt{x}-1}\) 2)\(\frac{x\sqrt{x}-2x}{2-\sqrt{x}}\) 3) \(\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\) 4) \(\frac{a\sqrt{b}-\sqrt{a}}{\sqrt{b}-b\sqrt{a}}\) 5) \(\frac{a-1}{\sqrt{a}+1}\) 6) \(\frac{4-x}{2\sqrt{x}-x}\) 7)\(\frac{a+1+2\sqrt{a}}{1+\sqrt{a}}\) 8)\(\frac{3\sqrt{x}-x}{3+2\sqrt{3x}-x}\) 9)\(\frac{y+12-4\sqrt{3y}}{y-12}\) 10)\(\frac{4\sqrt{x}-x-4}{x-4}\) 11)\(\frac{x+y-2\sqrt{xy}}{x\sqrt{y}-y\sqrt{x}}\)
RÚt gọn c)dfrac{1}{sqrt{1}+sqrt{2}}+dfrac{1}{sqrt{2}+sqrt{3}}+dfrac{1}{sqrt{3}+sqrt{4}}+...+dfrac{1}{sqrt{899}+sqrt{900}}
d)dfrac{4}{sqrt{7}-sqrt{3}}+dfrac{6}{3+sqrt{3}}+dfrac{sqrt{7}-7}{sqrt{7}-1}
e) dfrac{x+5-xsqrt{x-1}}{x-1-3sqrt{x-1}}
b) dfrac{sqrt{5}-sqrt{15}}{1-sqrt{3}}-sqrt{21+4sqrt{5}}
)) giúp câu nào được thì giúp ạ 3
Đọc tiếp
RÚt gọn c)\(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{899}+\sqrt{900}}\)
d)\(\dfrac{4}{\sqrt{7}-\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}+\dfrac{\sqrt{7}-7}{\sqrt{7}-1}\)
e) \(\dfrac{x+5-x\sqrt{x-1}}{x-1-3\sqrt{x-1}}\)
b) \(\dfrac{\sqrt{5}-\sqrt{15}}{1-\sqrt{3}}-\sqrt{21+4\sqrt{5}}\)
=)) giúp câu nào được thì giúp ạ <3
d) \(\dfrac{4}{\sqrt{7}-\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}+\dfrac{\sqrt{7}-7}{\sqrt{7}-1}\)
\(=\dfrac{4\left(\sqrt{7}+\sqrt{3}\right)}{\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)}+\dfrac{6\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}-\dfrac{7-\sqrt{7}}{\sqrt{7}-1}\)
\(=\dfrac{4\left(\sqrt{7}+\sqrt{3}\right)}{4}+\dfrac{6\left(3-\sqrt{3}\right)}{6}-\dfrac{\sqrt{7}\left(\sqrt{7}-1\right)}{\sqrt{7}-1}\)
\(=\sqrt{7}+\sqrt{3}+3-\sqrt{3}-\sqrt{7}=3\)
b) \(\dfrac{\sqrt{5}-\sqrt{15}}{1-\sqrt{3}}-\sqrt{21+4\sqrt{5}}=\dfrac{\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}-\sqrt{20+2\sqrt{20}+1}\)
\(=\sqrt{5}-\sqrt{\left(\sqrt{20}+1\right)^2}=\sqrt{5}-\left(\sqrt{20}+1\right)=\sqrt{5}-2\sqrt{5}-1=-1-\sqrt{5}\)
Đúng 0
Bình luận (0)
c) \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{899}+\sqrt{900}}\)
\(=\dfrac{\sqrt{1}-\sqrt{2}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{1}-\sqrt{2}\right)}+\dfrac{\sqrt{2}-\sqrt{3}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}+\dfrac{\sqrt{3}-\sqrt{4}}{\left(\sqrt{3}+\sqrt{4}\right)\left(\sqrt{3}-\sqrt{4}\right)}+...+\dfrac{\sqrt{899}-\sqrt{900}}{\left(\sqrt{899}+\sqrt{900}\right)\left(\sqrt{899}-\sqrt{900}\right)}\)\(=\dfrac{1-\sqrt{2}}{1-2}+\dfrac{\sqrt{2}-\sqrt{3}}{2-3}+\dfrac{\sqrt{3}-\sqrt{4}}{3-4}+...+\dfrac{\sqrt{899}-\sqrt{900}}{899-900}\)
\(=-\left(1-\sqrt{2}\right)-\left(\sqrt{2}-\sqrt{3}\right)-\left(\sqrt{3}-\sqrt{4}\right)+...-\left(\sqrt{899}-\sqrt{900}\right)\)
\(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}+...-\sqrt{899}+\sqrt{900}\)
\(=-1+\sqrt{900}\) \(=-1+30=29\)
b) \(\dfrac{\sqrt{5}-\sqrt{15}}{1-\sqrt{3}}-\sqrt{21+4\sqrt{5}}\)
\(=\dfrac{\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}-\sqrt{20+4\sqrt{5}+1}\)
\(=\sqrt{5}-\sqrt{\sqrt{20}+2\sqrt{20}+1}\)
\(=\sqrt{5}-\sqrt{\left(\sqrt{20}+1\right)^2}\)
\(=\sqrt{5}-\left(\sqrt{20}+1\right)\)
\(=\sqrt{5}-\sqrt{20}-1\)
\(=\sqrt{5}-2\sqrt{5}-1\)
\(=-\sqrt{5}-1\)
Đúng 0
Bình luận (0)