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a, \(\left(x-\frac{2}{3}\right)^2=\frac{5}{6}\Leftrightarrow\orbr{\begin{cases}x-\frac{2}{3}=\sqrt{\frac{5}{6}}\\x-\frac{2}{3}=-\sqrt{\frac{5}{6}}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{4+\sqrt{30}}{6}\\x=\frac{4-\sqrt{30}}{6}\end{cases}}}\)nghiệm xấu thế ?
b, \(\left(\frac{3}{4}-x\right)^3=\left(-8\right)\Leftrightarrow\frac{3}{4}-x=-2\Leftrightarrow x=\frac{11}{4}\)
Bài 1:
\(B=\frac{\frac{1}{2}+\frac{3}{4}-\frac{5}{6}}{\frac{1}{4}+\frac{3}{8}-\frac{5}{12}}+\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{8}}{\frac{1}{4}+\frac{1}{5}-\frac{1}{8}}\)\(=\frac{\frac{1}{2}+\frac{3}{4}-\frac{5}{6}}{\frac{1}{2}\left(\frac{1}{2}+\frac{3}{4}-\frac{5}{6}\right)}+\frac{3\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{8}\right)}{\frac{1}{4}+\frac{1}{5}-\frac{1}{8}}\)
\(=\frac{1}{\frac{1}{2}}+3\) \(=2+3\) \(=5\)
Vậy B=5
Bài 2:
a) x3 - 36x = 0
=> x(x2-36)=0
=> x(x2+6x-6x-36)=0
=> x[x(x+6)-6(x+6) ]=0
=> x(x+6)(x-6)=0
\(\Rightarrow\orbr{\begin{cases}^{x=0}x+6=0\\x-6=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}^{x=0}x=-6\\x=6\end{cases}}\)
Vậy x=0; x=-6; x=6
b) (x - y = 4 => x=4+y)
x−3y−2 =32
=>2(x-3) = 3(y-2)
=>2x-6= 3y-6
=>2x-3y=0
=>2(4+y)-3y=0
=>8+2y-3y=0
=>8-y=0
=>y=8 (thỏa mãn)
Do đó x=4+y=4+8=12 (thỏa mãn)
Vậy x=12 và y =8
B= 1/2 + 3/4 - 5/6/1/2(1.2 + 3/4 - 5/6) + 3(1/4+ 1/5 - 1/8)/ 1/4 1/5 - 1/8
B= 1/ 1/2 + 3
B= 2+3
B=5
B2:
a) x^3 - 36x = 0
x(x^2 - 36) = 0
=> x=0 hoặc x^2-36=0
=> x= 0 hoặc x^2=36
=> x=0 hoặc x= +- 6
câu 1 :
\(A=\frac{-7}{12}:\frac{49}{11}\cdot\frac{5}{121}-\frac{7}{6}\) \(B=\frac{1}{8}-\frac{8}{7}:8-3:\frac{3}{4}\cdot-2^3\)
\(A=\frac{-11}{84}\cdot\frac{5}{121}-\frac{7}{6}\) \(B=\frac{1}{8}-\frac{8}{7}:\frac{8}{1}-\frac{3}{1}:\frac{3}{4}\cdot\left(-2^3\right)\)
\(A=\frac{-5}{924}-\frac{7}{6}\) \(B=\frac{1}{8}-\frac{1}{7}-\left(-32\right)\)
\(A=\frac{-361}{308}\) \(B=\frac{-1}{56}-\left(-32\right)\)
\(B=\frac{1791}{56}\)
Câu 2 :
a)\(\frac{22}{7}:x=\frac{11}{7}\) b)\(\left(1-3x\right)\cdot\frac{4}{3}=-2^3\)
\(x=\frac{22}{7}:\frac{11}{7}\) \(\left(1-3x\right)\cdot\frac{4}{3}=-8\)
\(x=2\) \(\left(1-3x\right)=-8:\frac{4}{3}\)
\(\left(1-3x\right)=-6\)
\(3x=-6-1=7\)
\(3x=7:3=\frac{7}{3}\)
c ) bằng \(\frac{27}{5}\)nhé
\(2.THPT\)
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9\left(1-\frac{1}{100}\right)\)
\(A=9.\frac{99}{100}\)
\(A=\frac{891}{100}\)
\(B=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)
\(B=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)
\(B=\frac{1}{5}-\frac{1}{95}\)
\(B=\frac{18}{95}\)
\(D=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(D=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(D=\frac{1}{2}-\frac{1}{28}\)
\(D=\frac{13}{28}\)
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\(a)\) Ta có :
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(A=1-\frac{1}{2^{100}}< 1\)
Vậy \(A< 1\)
Chúc bạn học tốt ~
Bài làm :
a) Ta có :
\(\left(x-\frac{2}{3}\right)^2=\frac{5}{6}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{2}{3}=-\sqrt{\frac{5}{6}}\\x-\frac{2}{3}=\sqrt{\frac{5}{6}}\end{cases}}\)
\(\Leftrightarrow\frac{x=-\sqrt{\frac{5}{6}}+\frac{2}{3}}{x=\sqrt{\frac{5}{6}}+\frac{2}{3}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{4-\sqrt{30}}{6}\\x=\frac{4+\sqrt{30}}{6}\end{cases}}\)
b) Ta có :
\(\left(\frac{3}{4}-x\right)^2\ge0\text{ với mọi x}\)
\(\Rightarrow\left(\frac{3}{4}-x\right)^2\ne-8\)
=> Không tồn tại x
Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
a.
\(x-\frac{2}{3}=\pm\sqrt{\frac{5}{6}}\)
\(\orbr{\begin{cases}x-\frac{2}{3}=\sqrt{\frac{5}{6}}\\x-\frac{2}{3}=-\sqrt{\frac{5}{6}}\end{cases}}\)
\(\orbr{\begin{cases}x=\sqrt{\frac{5}{6}}+\frac{2}{3}\\x=\frac{2}{3}-\sqrt{\frac{5}{6}}\end{cases}}\)
b.
Ta có : \(\left(\frac{3}{4}-x\right)^2\ge0\forall x\)
\(\Rightarrow\) không có x