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Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
b: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{-3x-4y+5z+3-12-25}{-3\cdot2-4\cdot4+5\cdot6}=\dfrac{16}{8}=2\)
Do đó: x=5; y=5; z=17
\(a,\dfrac{x^3}{8}=\dfrac{y^3}{27}=\dfrac{z^3}{64}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}\)
Áp dụng t/c dtsbn:
\(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}=\dfrac{x^2+2y^2-3z^2}{4+18-48}=\dfrac{-650}{-26}=25\\ \Rightarrow\left\{{}\begin{matrix}x^2=100\\y^2=225\\z^2=400\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm10\\y=\pm15\\z=\pm20\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)\) có giá trị là hoán vị của \(\left(\pm10;\pm15;\pm20\right)\)
a: =>1/2x-3/4x=-5/6+7/3
=>-1/4x=14/6-5/6=3/2
=>x=-3/2*4=-6
b: =>4/5x-3/2x=1/2+6/5
=>-7/10x=17/10
=>x=-17/7
c: =>6/5x+6/20=6/5-1/3x
=>6/5x+1/3x=6/5-3/10=12/10-3/10=9/10
=>x=27/46
d: =>6x+3/2+4/5=1/2-2x
=>8x=1/2-3/2-4/5=-1-4/5=-9/5
=>x=-9/40
a, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{x+y}{3+5}=\dfrac{16}{8}=2\\ \Rightarrow\left\{{}\begin{matrix}x=6\\y=10\end{matrix}\right.\\ b,x:2=y:\left(-5\right)\Rightarrow\dfrac{x}{2}=\dfrac{y}{-5}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{2}=\dfrac{y}{-5}=\dfrac{x-y}{2-\left(-5\right)}=\dfrac{-7}{7}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-2\\y=5\end{matrix}\right.\)
a: =>x+5>0 và x-2<0
=>-5<x<2
=>x thuộc {-4;-3;...;1}
b: =>(x-5)(x+5)>0
=>x>5 hoặc x<-5
=>x thuộc Z\{-5;-4;-3;...;3;4;5}
c: =>(x+6)(x-7)>0
=>x>7 hoặc x<-6
`#3107.101107`
`1.`
`a,`
`(2x - 3)^2 = |3 - 2x|`
`=> (2x - 3)^2 = |2x - 3|`
`=>`\(\left[{}\begin{matrix}2x-3=\left(2x-3\right)^2\\2x-3=-\left(2x-3\right)^2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x-3-\left(2x-3\right)^2=0\\2x-3+\left(2x-3\right)^2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}\left(2x-3\right)\left(1-2x+3\right)=0\\\left(2x-3\right)\left(1+2x-3\right)=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x-3=0\\4-2x=0\\2x-2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\\x=1\end{matrix}\right.\)
Vậy, `x \in {3/2; 2; 1}`
`b,`
`(x - 1)^2 + (2x - 1)^2 = 0`
`=>`\(\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(2x-1\right)^2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x-1=0\\2x-1=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy, `x \in {1; 1/2}`
`c,`
`5 - x^2 = 1`
`=> x^2 = 4`
`=> x^2 = (+-2)^2`
`=> x = +-2`
Vậy, `x \in {-2; 2}`
`d,`
`x - 2\sqrt{x} = 0`
`=> x^2 - (2\sqrt{x})^2 = 0`
`=> x^2 - 4x = 0`
`=> x(x - 4) = 0`
`=>`\(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy, `x \in {0; 4}`
`g,`
`(x - 1) + 1/7 = 0`
`=> x - 1 + 1/7 = 0`
`=> x - 6/7 = 0`
`=> x = 6/7`
Vậy, `x = 6/7.`
a) \(\left|\dfrac{2}{7}\right|\) = \(\dfrac{2}{7}\)
b) \(\left|\dfrac{-5}{6}\right|\) = \(\dfrac{5}{6}\)
c) \(\left|4\dfrac{2}{3}\right|\) = \(4\dfrac{2}{3}\)
d) \(\left|-3,41\right|\) = \(3,41\)
\(a,\dfrac{3}{7}-x=\dfrac{1}{2}x-3\)
\(\Rightarrow-x-\dfrac{1}{2}x=-3-\dfrac{3}{7}\)
\(\Rightarrow-\dfrac{3}{2}x=-\dfrac{24}{7}\)
\(\Rightarrow x=-\dfrac{24}{7}:\left(-\dfrac{3}{2}\right)\)
\(\Rightarrow x=\dfrac{16}{7}\)
\(b,5x-\dfrac{2}{3}=\dfrac{5}{3}-2x\)
\(\Rightarrow5x+2x=\dfrac{5}{3}+\dfrac{2}{3}\)
\(\Rightarrow7x=\dfrac{7}{3}\)
\(\Rightarrow x=\dfrac{7}{3}:7\)
\(\Rightarrow x=\dfrac{1}{3}\)
#Toru
a: 3/7-x=1/2x-3
=>-3/2x=-3+3/7
=>-1/2x=-1+1/7=-6/7
=>1/2x=6/7
=>x=6/7*2=12/7
b: =>5x+2x=5/3+2/3
=>7x=7/3
=>x=1/3
a, 3(x - 2) = 5 - 2(x + 7)
=> 3x - 6 = 5 - 2x - 14
=> 3x - 6 = -9 - 2x
=> 3x + 2x = -9 + 5
=> 5x = -4
=> x = -4/5
4|x + 3| - 145 = 15
=> 4|x + 3| = 160
=> |x + 3| = 40
=> x + 3 = 40 hoặc x + 3 = -40
=> x = 37 hoặc x = -43
vậy_