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Bài 1:
\(4.\left(\frac{-1}{2}\right)^2-2.\left(\frac{-1}{2}\right)^2+3.\left(\frac{-1}{2}\right)+1\)
\(=4.\frac{1}{4}-2.\frac{1}{4}+3.\left(\frac{-1}{2}\right)+1\)
\(=1-\frac{1}{2}-\frac{3}{2}+1\)
\(=0\)
Bài 2:
a) \(\frac{37-x}{x+13}=\frac{3}{7}\)
\(\Rightarrow7\left(37-x\right)=3\left(x+13\right)\)
\(\Rightarrow259-7x=3x+39\)
\(\Rightarrow259-39=3x+7x\)
\(\Rightarrow220=10x\)
\(\Rightarrow x=22\)
d) \(\frac{3^2.3^8}{27^3}=3^x\)
\(\Rightarrow\frac{3^{10}}{\left(3^3\right)^3}=3^x\)
\(\frac{\Rightarrow3^{10}}{3^9}=3^x\)
\(\Rightarrow3=3^x\)
\(\Rightarrow x=1\)
Hok tốt nha^^
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{2}{3}\right|+\left|x+\frac{3}{4}\right|=4x\)
\(\Rightarrow x=x\)hoặc \(x=-x\)
---Nếu x = x
\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{2}{3}\right|+\left|x+\frac{3}{4}\right|=4x\)
\(x+\frac{1}{2}+x+\frac{2}{3}+x+\frac{3}{4}=4x\)
\(3x+\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}\right)=4x\)
\(\left(\frac{6}{12}+\frac{8}{12}+\frac{9}{12}\right)=4x-3x\)
\(\Rightarrow x=\frac{23}{12}\)
---Nếu x = -x
\(\Rightarrow\left|-x+\frac{1}{2}\right|+\left|-x+\frac{2}{3}\right|+\left|-x+\frac{3}{4}\right|=4x\)
\(-x+\frac{1}{2}+\left(-x\right)+\frac{2}{3}+\left(-x\right)+\frac{3}{4}=4x\)
\(-3x+\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}\right)=4x\)
\(\left(\frac{6}{12}+\frac{8}{12}+\frac{9}{12}\right)=4x+3x\)
\(\Rightarrow7x=\frac{23}{12}\)
\(\Rightarrow x=\frac{23}{12}:7\)
\(\Rightarrow x=\frac{23}{12}.\frac{1}{7}\)
\(\Rightarrow x=\frac{23}{84}\)
\(Vậyx=\frac{23}{12};x=\frac{23}{84}\)
a/\(\left(2-x\right)\times-3=\left(3x-1\right)\times4\)4
\(\Rightarrow-6+3x=12x-4\)
\(\Rightarrow-2=9x\)
\(\Rightarrow x=\frac{-2}{9}\)
bài b cx tương tự nha
ta có;\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}=\frac{a+b}{c+d}\)(THEO TÍNH CHẤT DÃY TỈ SỐ BẰNG NHAU)
\(\Rightarrowđpcm\)
a, 2/1/3:1/3=7/9:x b, x:1/3=12/99:15/90
7 = 7/9 : x x :1/3= 8/11
x = 7/9:7 x = 8/11 * 1/3
x = 1/9 x = 8/33
c, 0,15:x=3/1/3:2,25 d, 3/4:0,75=x:75/90
0,15:x =40/27 1 =x:75/90
x = 0,15:40/27 x = 1*75/90
x = 81/800 x = 75/90
e, x/-15=-60/x f, -2/x=-x/8
=>x*x=-15*(-60) => (-2)*8=x*-x
=>x2=900 => -16 = -x2
=>x2=302 hoặc x2=(-30)2 => 16=x2
=> x=30 hoặc x= -30 => x2=42 hoặc x2=(-4)2
=> x=4 hoặc x=-4
a)\(2\frac{1}{3}:\frac{1}{3}=\frac{7}{9}:x\)
\(\frac{7}{3}\times3=\frac{7}{9}:x\)
\(7=\frac{7}{9}:x\)
\(x=\frac{7}{9}:7\)
\(x=\frac{7}{9}\times\frac{1}{7}\)
\(x=\frac{1}{9}\)
\(a,\frac{3x+2}{5x+7}=\frac{3x-1}{5x-1}=\frac{\left(3x+2\right)-\left(3x-1\right)}{\left(5x+7\right)-\left(5x-1\right)}=\frac{3}{8};\frac{3x+2}{5x+7}=\frac{3}{8}\Leftrightarrow24x+16=15x+21\Leftrightarrow9x=5\Leftrightarrow x=\frac{5}{9}\) \(b,\frac{37-x}{x+13}=\frac{3}{7}\Leftrightarrow37.7-7x=3x+39\Leftrightarrow259-7x=3x+39\Leftrightarrow220-7x=3x\Leftrightarrow10x=220\Leftrightarrow x=22\) \(c,\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}=\frac{x+4}{2x+6}=\frac{\left(x+4\right)-\left(x+1\right)}{2x+6-\left(2x+1\right)}=\frac{3}{5};\frac{x+1}{2x+1}=\frac{3}{5}\Leftrightarrow5x+5=6x+3\Leftrightarrow x=2\) \(d,\frac{x-2}{x+2}=\frac{x+3}{x-4}=\frac{\left(x+3\right)-\left(x-2\right)}{\left(x-4\right)-\left(x+2\right)}=\frac{5}{-6};\frac{x-2}{x+2}=\frac{5}{-6}\Leftrightarrow6\left(2-x\right)=5x+10\Leftrightarrow2-6x=5x\Leftrightarrow x=\frac{2}{11}\) \(f,\frac{3x-5}{x}=\frac{9x}{3x+2}=\frac{9x-15}{3x}=\frac{9x-\left(9x-15\right)}{\left(3x+2\right)-3x}=\frac{15}{2};\frac{9x}{3x+2}=\frac{15}{2}\Leftrightarrow18x=45x+30\Leftrightarrow27x+30=0\Leftrightarrow x=\frac{-10}{9}\) \(e,\frac{x+2}{6}=\frac{5x-1}{5}\Leftrightarrow5\left(x+2\right)=6\left(5x-1\right)\Leftrightarrow5x+10=30x-6\Leftrightarrow10=25x-6\Leftrightarrow25x=16\Leftrightarrow x=\frac{16}{25}\)
a) Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{-4}=\frac{x-y-z}{2-3+4}=\frac{27}{3}=9\)
=> \(\hept{\begin{cases}\frac{x}{2}=9\\\frac{y}{4}=9\\\frac{z}{-4}=9\end{cases}}\) => \(\hept{\begin{cases}x=9.2=18\\y=9.3=27\\z=9.\left(-4\right)=-36\end{cases}}\)
Vậy ...
a, ÁP DỤNG DÃY TỈ SỐ BĂNG NHAU TA CÓ
\(\frac{x}{2}=\frac{y}{3}=\frac{x}{-4}=\frac{x-y-z}{2-3+4}=\frac{27}{3}=9\)
\(\Rightarrow\hept{\begin{cases}x=9.2=18\\y=9.3=27\\z=9.\left(-4\right)=-36\end{cases}}\)
\(A=\frac{99}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{99.100}\right)\)
\(A=\frac{99}{100}-\left(1-\frac{1}{100}\right)\)
\(A=\frac{99}{100}-\frac{99}{100}\)
\(A=\frac{99-99}{100}=0\)
Bài 2
\(\left(3x+5\right).\left(2x-4\right)=0\)
\(TH1:3x+5=0\)
\(3x=-5\)
\(x=-\frac{5}{3}\)
\(TH2:2x-4=0\)
\(2x=4\)
\(x=2\)
\(\left(x^2-1\right).\left(x+3\right)=0\)
\(\Rightarrow x^2-1=0\)
\(x^2=1\)
\(\Rightarrow x=1\)
\(x+3=0\)
\(x=-3\)
\(5x^2-\frac{1}{2}x=0\)
\(\Rightarrow5x^2-\frac{x}{2}=0\)
\(\Rightarrow5x^2=\frac{5x^2}{1}=\frac{5x^2.2}{2}\)
\(10x^2-x=x.\left(10x-1\right)\)
\(\frac{x.\left(10x-1\right)}{2}=0\)
\(\frac{x.\left(10x-1\right)}{2}.2=0.2\)
\(10x-1=0\)
\(x=\frac{1}{10}=0.100\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{10}=0.100\\x=0\end{cases}}\)
\(\frac{x}{4}-\frac{1}{2}=\frac{3}{4}\)
\(\frac{x}{4}=\frac{3}{4}+\frac{1}{2}\)
\(\frac{x}{4}=\frac{5}{4}\)
\(\Rightarrow x=5\)
\(\frac{1}{8}+\frac{7}{8}:x=\frac{3}{4}\)
\(\frac{7}{8}:x=\frac{3}{4}-\frac{1}{8}\)
\(x=\frac{7}{8}:\frac{5}{8}\)
\(x=\frac{56}{40}=\frac{28}{20}=\frac{14}{10}=\frac{7}{5}\)
a)\(\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{7}\right)^7\)
\(x=\left(\frac{3}{7}\right)^7\div\left(\frac{3}{7}\right)^5\)
\(x=\left(\frac{3}{7}\right)^2\)
\(x=\frac{9}{49}\)
Vậy...
b)\(\left(-\frac{1}{3}\right)^3.x=\left(\frac{1}{3}\right)^4\)
\(\left(-\frac{1}{3}\right)^3.x=\left(-\frac{1}{3}\right)^4\)
\(x=\left(-\frac{1}{3}\right)^4\div\left(\frac{-1}{3}\right)^3\)
\(x=-\frac{1}{3}\)
Vậy...
c)\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
=>\(x-\frac{1}{2}=\frac{1}{3}\)
\(x=\frac{1}{3}+\frac{1}{2}\)
\(x=\frac{5}{6}\)
Vậy...
d)\(\left(x+\frac{1}{4}\right)^4=\left(\frac{2}{3}\right)^4\)
=>\(x+\frac{1}{4}=\frac{2}{3}\)
\(x=\frac{2}{3}-\frac{1}{4}\)
\(x=\frac{5}{12}\)
Vậy...
Phù, mãi mới xong, tk cho mk nha bn
Ta có: \(\frac{x}{4}=\frac{27}{x}\Leftrightarrow x^2=108\Leftrightarrow x=\sqrt{108}\)
\(\left(3x-1\right)^3=\frac{-8}{27}\)
Mà \(\frac{-8}{27}=\left(\frac{-2}{3}\right)^3\Rightarrow\left(3x-1\right)^3=\left(\frac{-2}{3}\right)^3\)
\(\Rightarrow3x-1=\frac{-2}{3}\Rightarrow3x=\frac{1}{3}\Rightarrow x=\frac{1}{9}\)
Vậy x = 1/9
+) \(\frac{x}{4}=\frac{27}{x}\)
\(\Rightarrow x\times x=27\times4\)
\(\Rightarrow x^2=108\)
\(\Rightarrow x=\sqrt{108}\)
Vậy \(x=\sqrt{108}\)
+) \(\left(3x-1\right)^3=\frac{-8}{27}\)
\(\Rightarrow\left(3x-1\right)^3=\left(\frac{-2}{3}\right)^3\)
\(\Rightarrow3x-1=\frac{-2}{3}\)
\(\Rightarrow3x=\frac{-2}{3}+1\)
\(\Rightarrow3x=\frac{1}{3}\)
\(\Rightarrow x=\frac{1}{3}:3\)
\(\Rightarrow x=\frac{1}{9}\)
Vậy \(x=\frac{1}{9}\)
_Chúc bạn học tốt_