Cho A=\(\frac{\sqrt{\sqrt{3}+1}-\sqrt{\sqrt{3}-1}}{\sqrt{\sqrt{3}+1}+\sqrt{\sqrt{3}-1}}\)Hỏi A có phải là nghiệm của phương trình: \(\left(x^4+x+2\right)\left(x+2-\frac{\sqrt{6}-\sqrt{30}}{\sqrt{2}-\sqrt{10}}\right)=0\)
Những câu hỏi liên quan
1/Tính:
a)\(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
b)\(\left(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}\right).\left(3\sqrt{\frac{2}{3}}-\sqrt{12}-\sqrt{6}\right)\)
2/Giải phương trình:
a)\(\sqrt{x^2-2x+1}=7\)
b) \(\sqrt{4x-20}-3\sqrt{\frac{x-5}{9}}=\sqrt{1-x}\)
bài 1:
a:\(\sqrt{\left(\sqrt{3}-2\right)^2}\)+\(\sqrt{\left(1+\sqrt{3}\right)^2}\)
=\(\sqrt{3}-2+1+\sqrt{3}\)
=\(2\sqrt{3}-1\)
b; dài quá mink lười làm thông cảm
bài 2:
\(\sqrt{x^2-2x+1}=7\)
=>\(\sqrt{\left(x-1\right)^2}=7
\)
=>\(\orbr{\begin{cases}x-1=7\\x-1=-7\end{cases}}\)
=>\(\orbr{\begin{cases}x=8\\x=-6\end{cases}}\)
b: \(\sqrt{4x-20}-3\sqrt{\frac{x-5}{9}}=\sqrt{1-x}\)
=>\(\sqrt{4\left(x-5\right)}-9\sqrt{x-5}=\sqrt{1-x}\)
\(=2\sqrt{x-5}-9\sqrt{x-5}=\sqrt{1-x}\)
=>\(-7\sqrt{x-5}=\sqrt{1-x}\)
=\(-7.\left(x-5\right)=1-x\)
=>\(-7x+35=1-x\)
=>\(-7x+x=1-35\)
=>\(-6x=-34\)
=>\(x\approx5.667\)
mink sợ câu b bài 2 sai đó bạn
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1 a)\(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
= \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
= \(|2-\sqrt{3}|+|1+\sqrt{3}|\)
= \(2-\sqrt{3}+1+\sqrt{3}\)
= \(2+1\)= \(3\)
b) \(\left(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}\right)\cdot\left(3\sqrt{\frac{2}{3}}-\sqrt{12}-\sqrt{6}\right)\)
= \(\left(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{6}{3^2}}-4\sqrt{\frac{6}{2^2}}\right)\cdot\left(3\sqrt{\frac{6}{3^2}}-\sqrt{6}\sqrt{2}-\sqrt{6}\right)\)
= \(\left(\frac{3}{2}\sqrt{6}+\frac{2}{3}\sqrt{6}-\frac{4}{2}\sqrt{6}\right)\cdot\left(\frac{3}{3}\sqrt{6}-\sqrt{6}\cdot\sqrt{2}-\sqrt{6}\right)\)
= \(\left(\frac{3}{2}\sqrt{6}+\frac{2}{3}\sqrt{6}-2\sqrt{6}\right)\cdot\left(\sqrt{6}-\sqrt{6}\cdot\sqrt{2}-\sqrt{6}\right)\)
= \(\left(\sqrt{6}\left(\frac{3}{2}+\frac{2}{3}-2\right)\right)\cdot\left(\sqrt{6}\left(1-\sqrt{2}-1\right)\right)\)
= \(\sqrt{6}\frac{1}{6}\cdot\sqrt{6}\left(-\sqrt{2}\right)\)
= \(\sqrt{6}^2\left(\frac{-\sqrt{2}}{6}\right)\)
= \(6\frac{-\sqrt{2}}{6}\)=\(-\sqrt{2}\)
2 a) \(\sqrt{x^2-2x+1}=7\)
<=> \(\sqrt{x^2-2x\cdot1+1^2}=7\)
<=> \(\sqrt{\left(x-1\right)^2}=7\)
<=> \(|x-1|=7\)
Nếu \(x-1>=0\)=>\(x>=1\)
=> \(|x-1|=x-1\)
\(x-1=7\)<=>\(x=8\)(thỏa)
Nếu \(x-1< 0\)=>\(x< 1\)
=> \(|x-1|=-\left(x-1\right)=1-x\)
\(1-x=7\)<=>\(-x=6\)<=> \(x=-6\)(thỏa)
Vậy x=8 hoặc x=-6
b) \(\sqrt{4x-20}-3\sqrt{\frac{x-5}{9}}=\sqrt{1-x}\)
<=> \(\sqrt{4\left(x-5\right)}-3\frac{\sqrt{x-5}}{3}=\sqrt{1-x}\)
<=> \(2\sqrt{x-5}-\sqrt{x-5}=\sqrt{1-x}\)
<=> \(\sqrt{x-5}=\sqrt{1-x}\)
ĐK \(x-5>=0\)<=> \(x=5\)
\(1-x\)<=> \(-x=-1\)<=> \(x=1\)
Ta có \(\sqrt{x-5}=\sqrt{1-x}\)
<=> \(\left(\sqrt{x-5}\right)^2=\left(\sqrt{1-x}\right)^2\)
<=> \(x-5=1-x\)
<=> \(x-x=1+5\)
<=> \(0x=6\)(vô nghiệm)
Vậy phương trình vô nghiệm
Kết bạn với mình nha :)
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Giải phương trình:
\(\frac{2\left(x-\sqrt{3}\right)\left(x-\sqrt{2}\right)}{\left(1-\sqrt{2}\right)\left(1-\sqrt{3}\right)}+\frac{3\left(x-1\right)\left(x-\sqrt{3}\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}-\sqrt{3}\right)}+\frac{4\left(x-1\right)\left(x-\sqrt{2}\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}-\sqrt{2}\right)}=3x-1\)
Câu 1: Cho A frac{1}{sqrt{1}+sqrt{2}}+frac{1}{sqrt{2}+sqrt{3}}+...+frac{1}{sqrt{120}+sqrt{121}}Bfrac{1}{sqrt{1}}+frac{1}{sqrt{2}}+...+frac{1}{sqrt{35}} Chứng minh ABCâu 2: Tính Asqrt[3]{frac{X^3-3X+left(X^2-1right)sqrt{X^2-4}}{2}}+sqrt[3]{frac{X^3-3X+left(X^2-1right)sqrt{X^2-4}}{2}}Với xsqrt[3]{2017}Câu 3: Cho hai số thực x và y thoã mãn left(sqrt{X^2+1}+Xright)left(sqrt{Y^2+1}+Yright)1Tính x+yCâu 4: Trục căn thức mẫu số A frac{2}{2sqrt[3]{2}+2+sqrt[3]{4}}Câu 5 : Gọi a là nghiệm nguyên dương của...
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Câu 1: Cho A= \(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{120}+\sqrt{121}}\)B=\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{35}}\)
Chứng minh A<B
Câu 2: Tính A=\(\sqrt[3]{\frac{X^3-3X+\left(X^2-1\right)\sqrt{X^2-4}}{2}}+\sqrt[3]{\frac{X^3-3X+\left(X^2-1\right)\sqrt{X^2-4}}{2}}\)Với x=\(\sqrt[3]{2017}\)
Câu 3: Cho hai số thực x và y thoã mãn \(\left(\sqrt{X^2+1}+X\right)\left(\sqrt{Y^2+1}+Y\right)=1\)Tính x+y
Câu 4: Trục căn thức mẫu số A= \(\frac{2}{2\sqrt[3]{2}+2+\sqrt[3]{4}}\)
Câu 5 : Gọi a là nghiệm nguyên dương của Phương trình \(\sqrt{2}X^2+X-1=0\)Không giải pt tính
C=\(\frac{2a-3}{\sqrt{2\left(2a^4-2a+3\right)}+2a^2}\)
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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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
left(frac{x-3sqrt{x}}{x-9}-1right):left(frac{9-x}{x+sqrt{x}-6}-frac{sqrt{x}-3}{sqrt{x}-2}-frac{sqrt{x}-2}{sqrt{x}+3}right)left(frac{sqrt{a}+1}{sqrt{a}-1}-frac{sqrt{a}-1}{sqrt{a}+1}+4sqrt{a}right)left(sqrt{a}-frac{1}{sqrt{a}}right)left(frac{2sqrt{x}+x}{xsqrt{x}-1}-frac{1}{sqrt{x}+1}right):left(1-frac{sqrt{x}+2}{x+sqrt{x}+1}right)frac{sqrt{x}+2}{sqrt{x}+3}-frac{5}{x+sqrt{x}-6}+frac{1}{2-sqrt{x}}
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\(\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}-\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\)
\(\left(\frac{2\sqrt{x}+x}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}+1}\right):\left(1-\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(\frac{\sqrt{x}+2}{\sqrt{x}+3}-\frac{5}{x+\sqrt{x}-6}+\frac{1}{2-\sqrt{x}}\)
c) \(C=\frac{\left(2\sqrt{x}+x\right)\left(\sqrt{x}+1\right)-\left(x\sqrt{x}-1\right)}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{x+\sqrt{x}+1-\left(\sqrt{x}+2\right)}{x+\sqrt{x}+1}=\)
\(C=\frac{x\sqrt{x}+2x+x+2\sqrt{x}-x\sqrt{x}+1}{\left(\left(\sqrt{x}\right)^3-1\right)\left(\sqrt{x}+1\right)}\times\frac{x+\sqrt{x}+1}{x-1}=\)
\(C=\frac{3x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\times\frac{x+\sqrt{x}+1}{x-1}=\)
\(C=\frac{3x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\times\frac{1}{x-1}=\)
\(C=\frac{3x+2\sqrt{x}+1}{x-1}\times\frac{1}{x-1}=\frac{3x+2\sqrt{x}+1}{\left(x-1\right)^2}.\)
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d) ĐK: x>=0; x khác 4.
\(D=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-5-\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}.\)
\(D=\frac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(D=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)
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Xem thêm câu trả lời
Bài 1: Rút gọna. left(5-2sqrt{3}right)^2+left(5+2sqrt{3}right)^2b. left(sqrt{5}+sqrt{2}right)^2-left(2sqrt{5}+1right)left(2sqrt{5}-1right)-sqrt{40}c. left(sqrt{2}-1right)^2-frac{2}{3}sqrt{4}+frac{4sqrt{2}}{5}+sqrt{1frac{11}{15}}-sqrt{2}d. left(sqrt{6}-sqrt{18}+5sqrt{2}-frac{1}{2}sqrt{8}right)2sqrt{6}+2sqrt{3}e. left(2sqrt{3}-3sqrt{2}right)^2+6sqrt{6}+3sqrt{24}Bài 2: Rút gọnA left(frac{1}{x-sqrt{x}}+frac{1}{sqrt{x}-1}:frac{sqrt{x+1}}{x-2sqrt{x}+1}right)(x0 ; x khác 1)
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Bài 1: Rút gọn
a. \(\left(5-2\sqrt{3}\right)^2+\left(5+2\sqrt{3}\right)^2\)
b. \(\left(\sqrt{5}+\sqrt{2}\right)^2-\left(2\sqrt{5}+1\right)\left(2\sqrt{5}-1\right)-\sqrt{40}\)
c. \(\left(\sqrt{2}-1\right)^2-\frac{2}{3}\sqrt{4}+\frac{4\sqrt{2}}{5}+\sqrt{1\frac{11}{15}}-\sqrt{2}\)
d. \(\left(\sqrt{6}-\sqrt{18}+5\sqrt{2}-\frac{1}{2}\sqrt{8}\right)2\sqrt{6}+2\sqrt{3}\)
e. \(\left(2\sqrt{3}-3\sqrt{2}\right)^2+6\sqrt{6}+3\sqrt{24}\)
Bài 2: Rút gọn
A =\(\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}:\frac{\sqrt{x+1}}{x-2\sqrt{x}+1}\right)\)(x>0 ; x khác 1)
b, MA-Bfrac{sqrt{x}+2}{sqrt{x}+3}-left(frac{5}{x+sqrt{x}-6}+frac{1}{sqrt{x}-2}right)frac{sqrt{x}+2}{sqrt{x}+3}-frac{5}{x+sqrt{x}-6}-frac{1}{sqrt{x}-2}frac{left(sqrt{x}+2right)left(sqrt{x}-2right)}{x+sqrt{x}-6}-frac{5}{x+sqrt{x}-6}-frac{1left(sqrt{x}+3right)}{x+sqrt{x}-6}frac{x-4-5-sqrt{x}-3}{left(sqrt{x}+3right)left(sqrt{x}-2right)}frac{x-sqrt{x}-12}{left(sqrt{x}+3right)left(sqrt{x}-2right)}frac{x-4sqrt{x}+3sqrt{x}-12}{left(sqrt{x}+3right)left(sqrt{x}-2right)}frac{left(sqrt{x}-4right)left(sqrt{x...
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b, \(M=A-B=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\left(\frac{5}{x+\sqrt{x}-6}+\frac{1}{\sqrt{x}-2}\right)\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\frac{5}{x+\sqrt{x}-6}-\frac{1}{\sqrt{x}-2}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{x+\sqrt{x}-6}-\frac{5}{x+\sqrt{x}-6}-\frac{1\left(\sqrt{x}+3\right)}{x+\sqrt{x}-6}\)
\(=\frac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-4\sqrt{x}+3\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)\(=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)
bạn trung học hay tiểu học vậy
1)Giải phương trìnhsqrt{x}+sqrt[4]{20-x}42)Cho A frac{sqrt{1-sqrt{1-x^2}.left(sqrt{left(1+x^{ }right)}^3+sqrt{left(1-xright)^3}right)}}{2-sqrt{1-x^2}}Rút gọn và tìm x để A23)Cho Pfrac{sqrt{x}}{sqrt{xy}+sqrt{x}+2}+frac{sqrt{y}}{sqrt{yz}+sqrt{y}+1}+frac{2sqrt{x}}{sqrt{xz}+2sqrt{z}+2}Tìmsqrt{P}biết xyz1
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1)Giải phương trình
\(\sqrt{x}+\sqrt[4]{20-x}=4\)
2)Cho A= \(\frac{\sqrt{1-\sqrt{1-x^2}.\left(\sqrt{\left(1+x^{ }\right)}^3+\sqrt{\left(1-x\right)^3}\right)}}{2-\sqrt{1-x^2}}\)
Rút gọn và tìm x để A>=2
3)Cho P=\(\frac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+2}+\frac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\frac{2\sqrt{x}}{\sqrt{xz}+2\sqrt{z}+2}\)
Tìm\(\sqrt{P}\)
biết xyz=1