tìm x :
x : 12(x-1):3=4^3+2^3
giải chi tiét
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\(-\dfrac{1}{2}x+6< 0\Leftrightarrow-\dfrac{1}{2}x< -6\Leftrightarrow\cdot\dfrac{1}{2}x>6\Leftrightarrow x>12\)
(sai thì thoi nha)
\(-\dfrac{1}{2}x+6< 0\)
\(\Leftrightarrow-\dfrac{1}{2}x< -6\)
\(\Leftrightarrow x>\left(-6\right):\left(-\dfrac{1}{2}\right)\)
\(\Leftrightarrow x>12\)
--> Chọn A
\(x:\dfrac{1}{2}=\dfrac{3}{4}+\dfrac{5}{3}\\ \Rightarrow x:\dfrac{1}{2}=\dfrac{9}{12}+\dfrac{20}{12}\\ \Rightarrow x:\dfrac{1}{2}=\dfrac{29}{12}\\ \Rightarrow x=\dfrac{29}{12}\times\dfrac{1}{2}\\ \Rightarrow x=\dfrac{29}{24}\)
b: Ta có: \(\dfrac{12}{5}:x+\dfrac{4}{3}=3+\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{12}{5}:x=3-\dfrac{2}{3}=\dfrac{7}{3}\)
hay \(x=\dfrac{12}{5}:\dfrac{7}{3}=\dfrac{36}{35}\)
a. x + \(\dfrac{3}{7}\)= \(\dfrac{2}{5}:\dfrac{18}{25}=>x+\dfrac{3}{7}=\dfrac{2}{5}\)x\(\dfrac{35}{18}=>x+\dfrac{3}{7}=\dfrac{7}{9}\)
=> x = \(\dfrac{7}{9}-\dfrac{3}{7}=\dfrac{49}{63}-\dfrac{27}{63}=\dfrac{22}{63}\)
b. \(x\) x \(\dfrac{5}{9}\)= \(\dfrac{4}{5}-\dfrac{1}{3}\)
=> \(x\) x \(\dfrac{5}{9}\)= \(\dfrac{12}{15}-\dfrac{5}{15}=>x\) x \(\dfrac{5}{9}\)= \(\dfrac{7}{15}\)
=> x = \(\dfrac{7}{15}:\dfrac{5}{9}\)
=> x = \(\dfrac{21}{25}\)
\(a.x+\dfrac{3}{7}=\dfrac{2}{5}:\dfrac{18}{35}\\x+\dfrac{3}{7}=\dfrac{2}{5}\times\dfrac{35}{18} \\ x+\dfrac{3}{7}=\dfrac{7}{9}\\ x=\dfrac{7}{9}-\dfrac{3}{7}\\ x=\dfrac{22}{63}\)
\(b.x\times\dfrac{5}{9}=\dfrac{4}{5}-\dfrac{1}{3}\\x\times\dfrac{5}{9}=\dfrac{7}{15}\\ x=\dfrac{7}{15}:\dfrac{5}{9}\\ x= \dfrac{21}{25}\)
\(\Leftrightarrow\left(x+3\right)^2-48x^2=\left(x-3\right)^2\)
\(\Leftrightarrow48x^2=x^2+6x+9-x^2+6x-9\)
\(\Leftrightarrow48x^2-12x=0\)
=>12x(4x-1)=0
=>x=0(nhận) hoặc x=1/4(nhận)
a, Thay x = 3 và y = -6 vào bt ta đc
\(5.3-4.\left(-6\right)=15-\left(-24\right)=39\\ b,\\ 2.\left(-2\right)^2-5.4=8-20=\left(-12\right)\\ c,\\ 5.\left(-1\right)^2+3.\left(-1\right)-1=5+\left(-3\right)-1=1\)
a) Thay x=3; y=-6
\(5x-4y=5.3-4.\left(-6\right)=15+24=39\)
b) Thay x=-2; y=4
\(2x^4-5y=2.\left(-2\right)^4-5.4=32-20=12\)
c, Thay x=0
\(5x^2+3x-1=5.0+3.0-1=-1\)
+) x=-1
\(5x^2+3x-1=5.\left(-1\right)^2+3.\left(-1\right)-1=5-3-1=1\)
+) \(x=\dfrac{1}{3}\)
\(5x^2+3x-1=5.\left(\dfrac{1}{3}\right)^2+3.\dfrac{1}{3}-1\)
\(=\dfrac{5}{9}+1-1=\dfrac{5}{9}\)
=>3x+3+5x-5=3x^2-3
=>3x^2-3=8x-2
=>3x^2-8x-1=0
=>\(x=\dfrac{4\pm\sqrt{19}}{3}\)