giải giúp mik bt này nha,sáng mai mik phải nộp rùi.Thanks các bạn nhìu
Tìm x biết:
\(3\frac{1}{3}x+16\frac{3}{4}=-13,25\)
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\(\frac{1}{x^2+3}+\frac{1}{x^2+9x+18}+\frac{1}{x^2+15x+54}=\frac{1}{2}\left(27-\frac{1}{x+9}\right)\)
\(\Leftrightarrow\frac{3}{x\left(x+3\right)}+\frac{3}{\left(x+3\right)\left(x+6\right)}+\frac{3}{\left(x+6\right)\left(x+9\right)}=27-\frac{1}{x+9}\)
Mà
\(\frac{3}{x\left(x+3\right)}+\frac{3}{\left(x+3\right)\left(x+6\right)}+\frac{3}{\left(x+6\right)\left(x+9\right)}\)
\(=\frac{1}{x}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+9}\)
\(=\frac{1}{x}-\frac{1}{x+9}\)
\(\Rightarrow\frac{1}{x}=27\Rightarrow x=\frac{1}{27}\)
1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)
\(\left(4\frac{46}{65}+x\right).1\frac{1}{12}=5,75\)
\(\Rightarrow\frac{306}{65}+x.\frac{13}{12}=\frac{23}{4}\)
\(\Rightarrow\frac{51}{10}+\frac{13}{12}x=\frac{23}{4}\)
\(\Rightarrow306x=65x=345\)
\(\Rightarrow65x=39\)
\(\Rightarrow x=\frac{3}{5}\)
b, \(\frac{5}{4}-\left(\frac{3}{2}x+0,5\right)=1\frac{1}{4}\)
\(\Rightarrow\frac{5}{4}-\frac{3}{2}x-0,5=\frac{5}{4}\)
\(\Rightarrow\frac{5}{4}-\frac{3}{2}x-\frac{1}{2}=\frac{5}{4}\)
\(\Rightarrow\frac{3}{4}-\frac{3}{2}x=\frac{5}{4}\)
\(\Rightarrow3-6x=5\)
\(\Rightarrow-6x=2\)
\(\Rightarrow x=-\frac{1}{3}\)
Phần b) chị sai nhé ! Dấu [ ] là phần nguyên nâng cao của lớp 6 nhé.
Dễ mà
a, \(\left(4\frac{46}{65}+x\right).1\frac{1}{2}=5,75\)
\(\left(4\frac{46}{65}+x\right)=5,75:1\frac{1}{2}\)
\(\left(4\frac{46}{65}.x\right)=4\)
\(x=4:4\frac{46}{65}\)
\(x=\frac{130}{153}\)
b tương tự
a) \(=\frac{1}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.6}.\frac{6.6}{5.7}=\frac{6}{2.7}=\frac{3}{7}\)
B) \(=\frac{70}{11}+\frac{1}{9}-\frac{37}{11}-\frac{1}{9}=\left(\frac{70}{11}-\frac{37}{11}\right)+\left(\frac{1}{9}-\frac{1}{9}\right)=\frac{33}{11}+0=3\)
BÀI 2:
A) \(\Leftrightarrow\frac{7}{2}x-\frac{x}{2}+\frac{2x}{2}=\frac{7}{2}.\frac{5}{6}\)
\(\Leftrightarrow\frac{7x-x+2x}{2}=\frac{35}{12}\)
\(\Leftrightarrow\frac{8x}{2}=\frac{35}{12}\)
\(\Leftrightarrow8x.12=35.2\Leftrightarrow96x=70\Leftrightarrow x=\frac{70}{96}=\frac{35}{48}\)
b) \(\left(x-\frac{3}{1.2}\right)+\left(x-\frac{3}{2.3}\right)+...+\left(x-\frac{3}{99.100}\right)=1\)
\(x-\frac{3}{1.2}+x-\frac{3}{2.3}+....x+\frac{3}{99.100}=1\)
\(\Leftrightarrow\left(x+x+x+...+x\right)-3\left(\frac{1}{1.2}+\frac{1}{1.3}+....+\frac{1}{99.100}\right)=1\)
ngoặc 1 có 99 số hạng x
\(\Leftrightarrow99x-3\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)=1\)
\(\Leftrightarrow99x-3\left(1-\frac{1}{100}\right)=1\)
\(\Leftrightarrow99x-3.\frac{99}{100}=1\)
\(\Leftrightarrow99x=1+\frac{3.99}{100}\)
\(\Leftrightarrow99x=\frac{397}{100}\)
\(\Leftrightarrow x=\frac{397}{100.99}=\frac{397}{9900}\)
\(a,\frac{x+8}{3}+\frac{x+7}{2}=-\frac{x}{5}\)
\(\Leftrightarrow\frac{10\cdot\left(x+8\right)}{30}+\frac{15\left(x+7\right)}{30}=\frac{-6x}{30}\)
\(\rightarrow10x+80+15x+105=-6x\)
\(\Leftrightarrow31x+185=0\)
\(\Leftrightarrow x=-\frac{185}{31}\)
b,\(b,\frac{x-8}{3}+\frac{x-7}{4}=4+\frac{1-x}{5}\)
\(\Leftrightarrow\frac{20\left(x-8\right)}{60}+\frac{15\left(x-7\right)}{60}=\frac{240}{60}+\frac{12\left(1-x\right)}{60}\)
\(\rightarrow20x-160+15x-105=240+12-12x\)
\(\Leftrightarrow47x-517=0\)\(\Leftrightarrow x=11\)
\(\left(3\frac{1}{2}+2x\right).2\frac{2}{3}=5\frac{1}{3}\)
<=>\(\left(\frac{7}{2}+2x\right).\frac{8}{3}=\frac{16}{3}\)
<=>\(\frac{28}{3}+\frac{16x}{3}=\frac{16}{3}\)
<=>\(\frac{16x}{3}=\frac{-2}{3}\)
<=>\(16x=-2\)
<=>\(x=\frac{-1}{8}\)
vậy \(x=\frac{-1}{8}\)
b,\(\left|2x+3\right|=5\)
xét x<0,ta co: \(\left|2x+3\right|=5\)<=> \(-2x+3=5\)<=>\(-2x=2\)<=>\(x=-1\)(loại)
xét x>0,ta co:\(\left|2x+3\right|=5\)<=>\(2x+3=5\)<=>\(2x=2\)<=>\(x=1\)
c,\(\frac{x-2}{4}=\frac{5+x}{3}\)
<=>\(\frac{3x-6}{12}=\frac{20+4x}{12}\)
=>\(3x-6=20+4x\)
<=>\(3x-6-20-4x=0\)
<=>\(-x-26=0\)
<=>\(-x=26\)
<=>\(x=-26\)
kl:.......
\(\frac{10}{3}x+\frac{67}{4}=-\frac{53}{4}\)
<=> \(\frac{10}{3}x=-30\)
=> x = -9