Tifm x biết :
\(\left(\sqrt{2}\right)\)( x + 2 ) . ( x + 1 )
Những câu hỏi liên quan
Cho A=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}+\dfrac{6\sqrt{x}}{x-4}\right)\)và B=\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\) (x≥0;x≠4)
a)Rút gọn biểu thức A
b)Đặt M=A:B.Hãy so sánh M với 1
c)TIfm giá trị nhỏ nhất của M
-GIÚP MÌNH VỚI Ạ-
a: \(A=\left(\dfrac{\sqrt{x}}{x+2}+\dfrac{6\sqrt{x}}{x-4}\right)\cdot\dfrac{\sqrt{x}+2}{1}\)
\(=\dfrac{x-2\sqrt{x}+6\sqrt{x}}{x-4}\cdot\dfrac{\sqrt{x}+2}{1}=\dfrac{x+4\sqrt{x}}{\sqrt{x}-2}\)
b: \(M=A:B=\dfrac{x+4\sqrt{x}}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{x+4\sqrt{x}}{\sqrt{x}+1}\)
b: \(M-1=\dfrac{x+4\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{x+3\sqrt{x}-1}{\sqrt{x}+1}>0\)
=>M>1
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Bdfrac{sqrt{x}left(1-xright)^2}{1+sqrt{x}}divleft[left(dfrac{xsqrt{x}-1}{sqrt{x}-1}+sqrt{x}right)timesleft(dfrac{xsqrt{x}+1}{sqrt{x}+1}-sqrt{x}right)right]a) Rút gọn Bb) Tính B biết x4-2sqrt{3}c) Tìm x để B 2/3d) So sánh B với 1 e) Tìm x để B1f) Tìm xin Z để Bin Zg) Tìm GTNN của Bh) Tìm m để B.m 2 có nghiệm
Đọc tiếp
\(B=\dfrac{\sqrt{x}\left(1-x\right)^2}{1+\sqrt{x}}\div\left[\left(\dfrac{x\sqrt{x}-1}{\sqrt{x}-1}+\sqrt{x}\right)\times\left(\dfrac{x\sqrt{x}+1}{\sqrt{x}+1}-\sqrt{x}\right)\right]\)
a) Rút gọn B
b) Tính B biết \(x=4-2\sqrt{3}\)
c) Tìm x để B = 2/3
d) So sánh B với 1
e) Tìm x để B>1
f) Tìm \(x\in Z\) để \(B\in Z\)
g) Tìm GTNN của B
h) Tìm m để B.m = 2 có nghiệm
ĐKXĐ: x>=0; x<>1
a: \(B=\dfrac{\sqrt{x}\left(x-1\right)^2}{\sqrt{x}+1}:\left(\left(x+\sqrt{x}+1+\sqrt{x}\right)\left(x-\sqrt{x}+1-\sqrt{x}\right)\right)\)
\(=\dfrac{\sqrt{x}\left(x-1\right)^2}{\sqrt{x}+1}:\left[\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)^2\right]\)
\(=\dfrac{\sqrt{x}\left(x-1\right)^2}{\left(x-1\right)^2\cdot\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
b: Khi x=4-2căn 3=(căn 3-1)^2 thì \(B=\dfrac{\sqrt{3}-1}{\sqrt{3}-1+1}=\dfrac{\sqrt{3}-1}{\sqrt{3}}=\dfrac{3-\sqrt{3}}{3}\)
c: B=2/3
=>căn x/căn x+1=2/3
=>căn x=2
=>x=4
d: \(B-1=\dfrac{\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}=-\dfrac{1}{\sqrt{x}+1}< 0\)
=>B<1
e: B>1
=>-1/căn x+1>0
=>căn x+1<0(vô lý)
=>KO có x thỏa mãn
f: B nguyên khi căn x chia hết cho căn x+1
=>căn x+1-1 chia hết cho căn x+1
=>căn x+1=1 hoặc căn x+1=-1(loại)
=>căn x=0
=>x=0
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Pleft(frac{sqrt{x}-2}{x-1}-frac{sqrt{x}+2}{x+2sqrt{x}+1}right)frac{left(1-xright)^2}{2}Pleft(frac{sqrt{x}-2}{left(sqrt{x}-1right)left(sqrt{x}+1right)}-frac{sqrt{x}+2}{left(sqrt{x}+1right)^2}right)frac{left(1-xright)^2}{2}Pleft(frac{left(sqrt{x}-2right)left(sqrt{x}+1right)-left(sqrt{x}+2right)left(sqrt{x}-1right)}{left(x-1right)left(sqrt{x}+1right)}right)frac{left(x-1right)^2}{2}Pleft(frac{left(x-sqrt{x}-2right)-left(x+sqrt{x}-2right)}{left(x-1right)left(sqrt{x}+1right)}right)frac{left(x-1right)^...
Đọc tiếp
\(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\frac{\left(1-x\right)^2}{2}\)
\(P=\left(\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right)\frac{\left(1-x\right)^2}{2}\)
\(P=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}\right)\frac{\left(x-1\right)^2}{2}\)
\(P=\left(\frac{\left(x-\sqrt{x}-2\right)-\left(x+\sqrt{x}-2\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}\right)\frac{\left(x-1\right)^2}{2}\)
\(P=\frac{2\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}+1\right)}\frac{\left(x-1\right)^2}{2}\)
\(P=\frac{\sqrt{x}\left(x-1\right)}{\sqrt{x}+1}=\sqrt{x}\left(\sqrt{x}-1\right)=x-\sqrt{x}\)
frac{x+1}{2left(x-1right)}+frac{2}{2left(sqrt{x}+1right)}+frac{sqrt{x}}{sqrt{x}left(sqrt{x}-1right)}.frac{left(x+1right).sqrt{x}}{2sqrt{x}left(sqrt{x}-1right)left(sqrt{x}+1right)}+frac{2sqrt{x}left(sqrt{x}-1right)}{2sqrt{x}left(sqrt{x}-1right)left(sqrt{x}+1right)}+frac{2sqrt{x}left(sqrt{x}+1right)}{2sqrt{x}left(sqrt{x}-1right)left(sqrt{x}+1right)}.frac{xsqrt{x}+sqrt{x}}{2sqrt{x}left(sqrt{x}-1right)left(sqrt{x}+1right)}+frac{2x-2sqrt{x}}{2sqrt{x}left(sqrt{x}-1right)left(sqrt{x}+1right)}+frac{2x+2...
Đọc tiếp
\(=\frac{x+1}{2\left(x-1\right)}+\frac{2}{2\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\)
=\(\frac{\left(x+1\right).\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2\sqrt{x}\left(\sqrt{x}+1\right)}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\)
=\(\frac{x\sqrt{x}+\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2x-2\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2x+2\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\)
=\(\frac{x\sqrt{x}+4x+\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{\sqrt{x}\left(x+4\sqrt{x}+1\right)}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{\sqrt{x}\left(\sqrt{x}+1\right)^2}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\)
LƯU Ý: CAP NÀY CHỈ LÀ CAP NHÁP
Cho A=\(\dfrac{\sqrt{1-\sqrt{1-x^2}}\left[\sqrt{\left(1+x\right)^3}+\sqrt{\left(1-x\right)^3}\right]}{2-\sqrt{1-x^2}}\)
a)Rút gọn A
b)tìm x biết A≥ \(\dfrac{1}{2}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{1+x}=a\\\sqrt{1-x}=b\end{matrix}\right.\) \(\Rightarrow2=a^2+b^2\)
\(A=\dfrac{\sqrt{1-ab}\left(a^3+b^3\right)}{a^2+b^2-ab}=\dfrac{\sqrt{\dfrac{2}{2}-ab}\left(a+b\right)\left(a^2+b^2-ab\right)}{a^2+b^2-ab}\)
\(=\sqrt{\dfrac{a^2+b^2}{2}-ab}\left(a+b\right)=\left(a+b\right)\sqrt{\dfrac{\left(a-b\right)^2}{2}}=\dfrac{\left|a-b\right|\left(a+b\right)}{\sqrt{2}}\)
\(=\pm\dfrac{a^2-b^2}{\sqrt{2}}=\pm\dfrac{2x}{\sqrt{2}}=\pm\sqrt{2}x\)
b.
\(A\ge\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}\sqrt{2}x\ge\dfrac{1}{2}\left(x\ge0\right)\\-\sqrt{2}x\ge\dfrac{1}{2}\left(x\le0\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x\ge\dfrac{\sqrt{2}}{4}\\x\le-\dfrac{\sqrt{2}}{4}\end{matrix}\right.\)
Kết hợp ĐKXĐ \(\Rightarrow\left[{}\begin{matrix}\dfrac{\sqrt{2}}{4}\le x\le1\\-1\le x\le-\dfrac{\sqrt{2}}{4}\end{matrix}\right.\)
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14 tháng 10 2021 lúc 9:13
Giúp với
1) Thu gọn A
\(A=\left(\sqrt{x}-\dfrac{x+2}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}-4}{1-x}\right)\)
2) Tính A biết \(x=\left(\dfrac{2-\sqrt{5}}{2+\sqrt{5}}-\dfrac{2+\sqrt{5}}{2-\sqrt{5}}\right):\sqrt{20}\)
\(A=\dfrac{x+\sqrt{x}-x-2}{\sqrt{x}+1}:\dfrac{x-\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\\ A=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
\(x=\dfrac{9-4\sqrt{5}-9-4\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}:2\sqrt{5}=\dfrac{-8\sqrt{5}}{-2\sqrt{5}}=4\\ \Leftrightarrow\sqrt{x}=2\\ \Leftrightarrow A=\dfrac{2-1}{2+2}=\dfrac{1}{4}\)
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21 tháng 6 2021 lúc 16:38
Cho A=\(\dfrac{\sqrt{1-\sqrt{1-x^2}}.\left[\sqrt{\left(1+x\right)^3}+\sqrt{\left(1-x\right)^3}\right]}{2-\sqrt{1-x^2}}\)
a) Rút gọn A
b) Tìm x biết A\(\ge\) \(\dfrac{1}{2}\)
ĐKXĐ: \(-1\le x\le1\)
Đặt \(\left\{{}\begin{matrix}\sqrt{1-x}=a\\\sqrt{1+x}=b\end{matrix}\right.\) \(\Rightarrow a^2+b^2=2\) ta được:
\(A=\dfrac{\sqrt{1-ab}\left(a^3+b^3\right)}{2-ab}=\dfrac{\sqrt{\dfrac{a^2+b^2}{2}-ab}\left(a+b\right)\left(a^2+b^2-ab\right)}{a^2+b^2-ab}\)
\(=\sqrt{\dfrac{a^2+b^2-2ab}{2}}\left(a+b\right)=\dfrac{\left|a-b\right|\left(a+b\right)}{\sqrt{2}}\)
\(=\dfrac{\left|\sqrt{1-x}-\sqrt{1+x}\right|\left(\sqrt{1-x}+\sqrt{1+x}\right)}{\sqrt{2}}\)
- Với \(-1\le x\le0\Rightarrow A=\dfrac{\left(\sqrt{1-x}-\sqrt{1+x}\right)\left(\sqrt{1-x}+\sqrt{1+x}\right)}{\sqrt{2}}=-\sqrt{2}x\)
- Với \(0\le x\le1\Rightarrow A=\dfrac{\left(\sqrt{1+x}-\sqrt{1-x}\right)\left(\sqrt{1+x}+\sqrt{1-x}\right)}{\sqrt{2}}=\sqrt{2}x\)
b.
TH1: \(\left\{{}\begin{matrix}-1\le x\le0\\-\sqrt{2}x\ge\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow-1\le x\le-\dfrac{1}{2\sqrt{2}}\)
TH2: \(\left\{{}\begin{matrix}0\le x\le1\\\sqrt{2}x\ge\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow\dfrac{1}{2\sqrt{x}}\le x\le1\)
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Bleft(frac{xsqrt{x}+x+sqrt{x}}{xsqrt{x}-1}-frac{sqrt{x}+3}{1-sqrt{x}}right).frac{x-1}{2x+sqrt{x}-1} ĐKXĐ: ...frac{left(xsqrt{x}+x+sqrt{x}right)left(1-sqrt{x}right)-left(sqrt{x}+3right)left(xsqrt{x}-1right)}{left(xsqrt{x}-1right)left(1-sqrt{x}right)}.frac{x-1}{2x+2sqrt{x}-sqrt{x}-1}frac{xsqrt{x}+x+sqrt{x}-x^2-xsqrt{x}-x-x^2+sqrt{x}-3xsqrt{x}+3}{left(xsqrt{x}-1right)left(1-sqrt{x}right)}.frac{x-1}{2sqrt{x}left(sqrt{x}+1right)-left(sqrt{x}+1right)}frac{-3xsqrt{x}+2sqrt{x}-2x^2+3}{left(xsqrt{x}-1ri...
Đọc tiếp
\(B=\left(\frac{x\sqrt{x}+x+\sqrt{x}}{x\sqrt{x}-1}-\frac{\sqrt{x}+3}{1-\sqrt{x}}\right).\frac{x-1}{2x+\sqrt{x}-1}\) ĐKXĐ: ...
\(=\frac{\left(x\sqrt{x}+x+\sqrt{x}\right)\left(1-\sqrt{x}\right)-\left(\sqrt{x}+3\right)\left(x\sqrt{x}-1\right)}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{2x+2\sqrt{x}-\sqrt{x}-1}\)
\(=\frac{x\sqrt{x}+x+\sqrt{x}-x^2-x\sqrt{x}-x-x^2+\sqrt{x}-3x\sqrt{x}+3}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{2\sqrt{x}\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}\)
\(=\frac{-3x\sqrt{x}+2\sqrt{x}-2x^2+3}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{3-3x\sqrt{x}+2\sqrt{x}-2x^2}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{3\left(1-x\sqrt{x}\right)+2\sqrt{x}\left(1-x\sqrt{x}\right)}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(2\sqrt{x}+3\right)\left(1-x\sqrt{x}\right)}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-2\sqrt{x}-3}{1-\sqrt{x}}.\frac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-2\sqrt{x}-3}{1-\sqrt{x}}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2\sqrt{x}-1}\)
\(=\frac{2\sqrt{x}+3}{2\sqrt{x}-1}\)
Pleft(frac{sqrt{x}+2}{sqrt{x}+1}-frac{x-sqrt{x}-3}{x-sqrt{x}-2}right):left(frac{x-sqrt{x}}{x-sqrt{x}-2}+frac{2}{sqrt{x}-2}right)frac{left(sqrt{x}+2right)left(sqrt{x}-2right)-x+sqrt{x}+3}{left(sqrt{x}+1right)left(sqrt{x}-2right)}:frac{x-sqrt{x}+2left(sqrt{x}+1right)}{left(sqrt{x}+1right)left(sqrt{x}-2right)}frac{x-4-x+3+sqrt{x}}{left(sqrt{x}+1right)left(sqrt{x}-2right)}.frac{left(sqrt{x}+1right)left(sqrt{x}-2right)}{x-sqrt{x}+2sqrt{x}+2}frac{sqrt{x}-1}{x+sqrt{x}+2}
Đọc tiếp
\(P=\left(\frac{\sqrt{x}+2}{\sqrt{x}+1}-\frac{x-\sqrt{x}-3}{x-\sqrt{x}-2}\right):\left(\frac{x-\sqrt{x}}{x-\sqrt{x}-2}+\frac{2}{\sqrt{x}-2}\right)\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-x+\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}:\frac{x-\sqrt{x}+2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-4-x+3+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{x-\sqrt{x}+2\sqrt{x}+2}\)
\(=\frac{\sqrt{x}-1}{x+\sqrt{x}+2}\)
#)Hỏi j đi bn, bn ph hỏi cái j chứ làm lun rùi còn để cộng đồng ngắm ak ???
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Bó cả tay lẫn chân !!! Bất lực như gặp cực hình !
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Chắc là bạn ấy hỏi bạn ấy làm có đúng ko ha gì đó ?
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