\(\frac{-150.\left(1352-41\right)}{\left(1352-41\right).\left(150-15\right)}< x< \frac{\left(2400:48\right)-250}{350-\left(3600:12\right)}\)
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a)
\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
= \(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
= \(\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)
= \(\frac{1}{3}+\left(-1\right)+1\)
= \(\frac{1}{3}+0=\frac{1}{3}\)
b)
\(\frac{38}{45}-\left(\frac{8}{45}-\frac{17}{51}-\frac{3}{11}\right)\)
= \(\frac{38}{45}-\frac{8}{45}+\frac{17}{51}+\frac{3}{11}\)
= \(\left(\frac{38}{45}-\frac{8}{45}\right)+\left(\frac{17}{51}+\frac{3}{11}\right)\)
= \(\frac{2}{3}+\frac{20}{33}\)
= \(\frac{14}{11}\)
b) Ta có: \(x^3+4x+5=0\)
\(\Leftrightarrow x^3-x+5x+5=0\)
\(\Leftrightarrow x\left(x^2-1\right)+5\left(x+1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x-1\right)+5\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+5\right)=0\)
mà \(x^2-x+5>0\forall x\)
nên x+1=0
hay x=-1
Vậy: S={-1}
a)x2-(x+3)(3x+1)=9
⇔(x-3)(x+3)-(x+3)(3x+1)=0
⇔x+3=0 hoặc 3x+1=0
1.x+3=0 ⇔x=-3
2.3x+1=0⇔x=-1/3
phương trình có 2 nghiệm x=-3 và x=-1/3
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}\)
\(A=\frac{1.2.3...99}{2.3.4...100}\)
\(A=\frac{1}{100}\)
\(B=1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{72}\)
\(B=1+1+...+1+\left(\frac{1}{12}+\frac{1}{20}+...+\frac{1}{72}\right)\)
\(B=5.1+\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}\right)\)
\(B=5+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{8}-\frac{1}{9}\right)\)
\(B=5+\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(B=5+\frac{2}{9}=\frac{47}{9}\)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{99}{100}\)
\(=\frac{1.2.3.4....99}{2.3.4.5...100}\)
\(=\frac{1}{100}\)
a) 31/23 - ( 7/32 + 8/22)
= 31/23 - 7/32 + 8/23
= ( 31/23 + 8/23 ) - 7/32
= 32/22 - 7/32
= 39/32
Ccá ý khác làm tương tự