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1, \(=\frac{3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{11}+\frac{1}{13}\right)}{7\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{11}+\frac{1}{13}\right)}=\frac{3}{7}\)
2, a, \(\Leftrightarrow\left(3x-2\right)^{10}-\left(3x-2\right)^6=0\)
\(\Leftrightarrow\left(3x-2\right)^6\left[\left(3x-2\right)^4-1\right]=0\)
TH1: (3x-2)^6=0 <=> 3x-2=0 <=> x=2/3
TH2: (3x-2)^4-1=0 <=> (3x-2)^4=1
<=> 3x-2 = 1 hoặc 3x-2=-1
<=>x=1 hoặc x=-1/3
Vậy x=2/3 hoặc x=1 hoặc x=-1/3
b, \(\Leftrightarrow\orbr{\begin{cases}2x^2-13=-5\\2x^2-13=5\end{cases}\Leftrightarrow\orbr{\begin{cases}2x^2=8\\2x^2=18\end{cases}\Leftrightarrow}\orbr{\begin{cases}x^2=4\\x^2=9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\pm2\\x=\pm3\end{cases}}}\)
a) Ta có: \(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}\)
\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(5x+7\right)\left(3x-1\right)\)
\(\Leftrightarrow3x\left(5x+1\right)+2\left(5x+1\right)=5x\left(3x-1\right)+7\left(3x-1\right)\)
\(\Leftrightarrow15x^2+3x+10x+2=15x^2-5x+21x-7\)
\(\Leftrightarrow15x^2-15x^2+3x+10x+5x-21x=-7-2\)
\(\Leftrightarrow-3x=-9\)
\(\Leftrightarrow x=3\)
Vậy x = 3
b) Ta có: \(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\Leftrightarrow\left(x+1\right)\left(x+3\right)=\left(2x+1\right)\left(0,5x+2\right)\)
\(\Leftrightarrow x\left(x+3\right)+\left(x+3\right)=2x\left(0,5x+2\right)+\left(0,5x+2\right)\)
\(\Leftrightarrow x^2+3x+x+3=x^2+4x+0,5x+2\)
\(\Leftrightarrow x^2-x^2+3x+x-4x-0,5x=2-3\)
\(\Leftrightarrow-0,5x=-1\Leftrightarrow x=2\)
Vậy x = 2
Bài 4
x/2=y/3 va x.y=54
bài giải
Đặt x/2= y/3=k
=>x=2k,y=3k
=>2k.3k=54
6.k^2=54
=>k^2=54:6
=>k^2=9
=>k=3 hoặc k=-3
Với k=3 thĩ=6; y=9
Với k=-3 thì x=-6; y=-9
Vậy các cặp (x,y) thỏa mản (6,9):(-6<-9)
Nếu sai thi bảo tớ nhé
a) \(\frac{a}{4}=\frac{b}{6}\Rightarrow\frac{a}{20}=\frac{b}{30}\)
\(\frac{b}{5}=\frac{c}{8}\Rightarrow\frac{b}{30}=\frac{c}{48}\)
=> \(\frac{a}{20}=\frac{b}{30}=\frac{c}{48}\)
Áp dubgj tc của dãy tỉ số bằng nahu at có:
\(\frac{a}{20}=\frac{b}{30}=\frac{c}{48}=\frac{5a-3b-3c}{20\cdot5-30\cdot3-48\cdot3}=\frac{-536}{-134}=4\)
=> \(\begin{cases}a=80\\b=120\\c=192\end{cases}\)
b)Có: \(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}\)
=> \(\frac{a^2}{4}=\frac{b^2}{9}=\frac{c^2}{16}\)
Áp dụng tc của dãy tie số bằng nhau ta có:
\(\frac{a^2}{4}=\frac{b^2}{9}=\frac{c^2}{16}=\frac{a^2+3b^2-2c^2}{4+3\cdot9-2\cdot16}=\frac{-16}{-1}=16\)
=> \(\begin{cases}a=8;s=-8\\b=12;b=-12\\c=16;x=-16\end{cases}\)
Vậy (x;y;z) thỏa mãn là \(\left(8;12;16\right);\left(-8;-12;-16\right)\)
a)\(\left(\frac{1}{3}.x\right):\frac{2}{3}=\frac{7}{4}.\frac{5}{2}\)
\(\left(\frac{1}{3}.x\right):\frac{2}{3}=\frac{35}{8}\)
\(\frac{1}{3}.x=\frac{35}{8}.\frac{2}{3}\)
\(\frac{1}{3}.x=\frac{35}{12}\)
\(x=\frac{35}{12}:\frac{1}{3}\)
\(x=\frac{35}{12}.\frac{3}{1}\)
\(x=\frac{35}{4}\)
b)\(\frac{4,5}{0,3}=\frac{2,25}{0,1.x}\)
\(4,5.\left(0,1.x\right)=2,25.0,3\)
\(4,5.\left(0,1.x\right)=0,675\)
\(0,1.x=0,675:4,5\)
\(0,1.x=0,15\)
\(x=0,15:0,1\)
\(x=1.5\)
c)\(\frac{8}{\frac{1}{4}.x}=\frac{2}{0,02}\)
\(\left(\frac{1}{4}.x\right).2=8.0,02\)
\(\left(\frac{1}{4}.x\right).2=0,16\)
\(\frac{1}{4}.x=0,16:2\)
\(\frac{1}{4}.x=0,08\)
\(x=0,08:\frac{1}{4}\)
\(x=0,32\)
d)\(\frac{3}{\frac{9}{4}}=\frac{\frac{3}{4}}{6.x}\)
\(3.\left(6.x\right)=\frac{3}{4}.\frac{9}{4}\)
\(3.\left(6.x\right)=\frac{27}{16}\)
\(6.x=\frac{27}{16}:3\)
\(6.x=\frac{9}{16}\)
\(x=\frac{9}{16}:6\)
\(x=\frac{3}{32}\)
HỌC TỐT ^^
a. Vì \(\left|4-3x\right|\ge0\forall x\)\(\Rightarrow A=-\left|4-3x\right|+\frac{1}{2}\le\frac{1}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow-\left|4-3x\right|=0\Leftrightarrow3x=4\Leftrightarrow x=\frac{4}{3}\)
Vậy maxA = 1/2 <=> x = 4/3
b. Vì \(\left|x+0,1\right|\ge0\forall x\)\(\Rightarrow B=-\frac{7}{3}-\left|x+0,1\right|\le-\frac{7}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow-\left|x+0,1\right|=0\Leftrightarrow x=-0,1\)
Vậy maxB = - 7/3 <=> x = - 0,1