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\(\frac{5.18-10.27+15.36}{10.36-20.54+30.72}\)
\(=\frac{5.18-10.27+15.36}{5.2.18.2-10.2.27.2+15.2.36.2}\)
\(=\frac{5.18-10.27+15.36}{5.8.2.2-10.27.2.2+15.36.2.2}\)
\(=\frac{1}{2.2-2.2+2.2}\)
\(=\frac{1}{2.2}=\frac{1}{4}\)
\(\frac{5.18-10.27+15.36}{10.36-20.54+30.72}\)
\(=\frac{5.18-\left(5.2\right).27+\left(5.3\right).36}{\left(5.2\right).36-\left(5.4\right).54+\left(5.6\right).72}\)
\(=\frac{5.18-5.\left(2.27\right)+5.\left(3.36\right)}{5.\left(2.36\right)-5.\left(4.54\right)+5.\left(6.72\right)}\)
\(=\frac{5.18-5.54+5.108}{5.72-5.216+5.432}\)
\(=\frac{5.\left(18-54+108\right)}{5.\left(72-216+432\right)}\)
\(=\frac{5.\left(-36+108\right)}{5.\left(-144+432\right)}\)
\(=\frac{5.72}{5.288}\)
\(=\frac{5.72}{5..4.72}\)
\(=\frac{1}{4}\)
a) \(\frac{-3}{8}\)x \(16\frac{8}{7}\)- 0.375 x \(7\frac{9}{17}\)
=> \(\frac{-3}{8}\)x \(\frac{120}{7}\)- \(\frac{3}{8}\)x \(\frac{128}{17}\)= \(\frac{-3}{8}\)x ( \(\frac{120}{7}\)+\(\frac{128}{17}\))
=> = \(\frac{-3}{8}\)X \(\frac{2936}{119}\)=\(\frac{-1101}{119}\). ko thuận tiện đc nữa đâu nha ! tích đi mấy bạn !
\(\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{9999}{10000}\)
\(=\frac{1\cdot3}{2^2}\cdot\frac{2\cdot4}{3^2}\cdot\frac{3\cdot5}{4^2}\cdot...\cdot\frac{99\cdot101}{100^2}\)
\(=\frac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(99\cdot101\right)}{2^2\cdot3^2\cdot4^2\cdot...\cdot100^2}\)
\(=\frac{\left(1\cdot2\cdot3\cdot...\cdot99\right)\left(3\cdot4\cdot5\cdot101\right)}{\left(2\cdot3\cdot4\cdot...\cdot100\right)\left(2\cdot3\cdot4\cdot...\cdot100\right)}\)
\(=2\cdot101=202\)
cau a dau nhi cuoi cung k phai j dau nha ! mk an lom !
\(a,\)\(\left|x+5\right|=\frac{1}{7}-\left|\frac{4}{3}-\frac{1}{6}\right|\)
\(\Leftrightarrow\left|x+5\right|=\frac{1}{7}-\frac{7}{6}\)
\(\Leftrightarrow\left|x+5\right|=\frac{-43}{42}\)
ta có |x+5| \(\ge\)0 \(\forall x\)
Mà \(-\frac{43}{42}< 0\)nên ko có giá trị x thoả mãn
b,
\(\left|x+\frac{2}{3}\right|=\frac{1}{2}-\left(\frac{1}{4}+\frac{2}{3}\right)\)
\(\Leftrightarrow\left|x+\frac{2}{3}\right|=\frac{11}{12}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{2}{3}=\frac{11}{12}\forall x\ge-\frac{2}{3}\\-x-\frac{2}{3}=\frac{11}{12}\forall< -\frac{2}{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=-\frac{19}{12}\end{cases}}\)(thoả mãn đk)
bạn tính từng ngoặc rồi dùng chiệt tiêu của phép nhân phân số nhé
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right).....\left(\frac{1}{1999}-1\right)=-\frac{1}{2}.\left(-\frac{2}{3}\right).\left(-\frac{3}{4}\right).....\left(-\frac{1998}{1999}\right)=-\frac{1}{1999}\)
Có sai đề ko bạn phải là a.x=b.y=c.z chứ
Ta có a.x=b.y=c.z
=> \(x:\frac{1}{a}=y:\frac{1}{b}=z:\frac{1}{c}\)
\(\Rightarrow\frac{x}{\frac{1}{a}}=\frac{y}{\frac{1}{b}}=\frac{z}{\frac{1}{c}}=k\)
\(\Rightarrow\hept{\begin{cases}x=\frac{k}{a}\\y=\frac{k}{b}\\z=\frac{k}{c}\end{cases}}\)
Mà x.y.z =\(\frac{8}{abc}\)=>\(\frac{k}{a}.\frac{k}{b}.\frac{k}{c}=\frac{k^3}{abc}=\frac{8}{abc}\)
=>k\(^3\)=8
\(\Rightarrow\)k=2
\(\Rightarrow x=\frac{2}{a};y=\frac{2}{b};z=\frac{2}{c}\)
Học tốt
Đặt \(\frac{x}{4}=\frac{y}{3}=k\Rightarrow\hept{\begin{cases}x=4k\\y=3k\end{cases}}\)
xy = 12
<=> 4k.3k = 12
<=> 12k2 = 12
<=> k2 = 1
<=> k = ±1
Với k = 1 => x = 4 ; y = 3
Với k = -1 => x = -4 ; y = -3
\(\frac{5.18-10.27+15.36}{10.36-20.54+30.72}\)
\(=\frac{5.18-5.2.27+5.3.36}{10.36-10.2.54+10.3.72}\)
\(=\frac{5.18-5.54+5.108}{10.36-10.108+10.216}\)
\(=\frac{5.\left(18-54+108\right)}{10.\left(36-108+216\right)}\)
\(=\frac{5.72}{10.144}\)
\(=\frac{1.1}{2.2}\)
\(=\frac{1}{4}\)
Học tốt ❤⚽
\(=\frac{90-90.3+90.6}{90.4-90.12+90.24}=\frac{90.\left(1-3+6\right)}{90\left(4-12+24\right)}=\frac{4}{16}=\frac{1}{4}\)