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a, \(\dfrac{x}{2}+\dfrac{3x}{5}=-\dfrac{3}{2}\Rightarrow5x+6x=-15\Leftrightarrow x=-\dfrac{15}{11}\)
b, TH1 : \(\dfrac{2}{3}x-\dfrac{4}{7}=0\Leftrightarrow x=\dfrac{6}{7}\);TH2 : \(\dfrac{1}{2}-\dfrac{3}{7x}=0\Rightarrow7x-6=0\Leftrightarrow x=\dfrac{6}{7}\)
c, TH1 : \(\dfrac{4}{5}-2x=0\Leftrightarrow x=\dfrac{4}{5}:2=\dfrac{2}{5}\)
TH2 : \(\dfrac{1}{3}+\dfrac{3}{5x}=0\Rightarrow5x+9=0\Leftrightarrow x=-\dfrac{9}{5}\)
`#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.`
\(a,\dfrac{3}{2}\cdot x-1=\dfrac{1}{2}x-\dfrac{3}{5}\)
\(\Rightarrow\dfrac{3}{2}x-\dfrac{1}{2}x=-\dfrac{3}{5}+1\)
\(\Rightarrow\left(\dfrac{3}{2}-\dfrac{1}{2}\right)x=-\dfrac{3}{5}+\dfrac{5}{5}\)
\(\Rightarrow x=\dfrac{2}{5}\)
\(b,\dfrac{1}{2}x+\dfrac{1}{2}\left(x-2\right)=\dfrac{3}{4}-2x\)
\(\Rightarrow\dfrac{1}{2}x+\dfrac{1}{2}x+2x-1=\dfrac{3}{4}\)
\(\Rightarrow\left(\dfrac{1}{2}+\dfrac{1}{2}+2\right)x=\dfrac{3}{4}+1\)
\(\Rightarrow3x=\dfrac{7}{4}\)
\(\Rightarrow x=\dfrac{7}{4}:3\)
\(\Rightarrow x=\dfrac{7}{12}\)
\(c,\left(x-\dfrac{1}{2}\right)-\dfrac{1}{4}=0\)
\(\Rightarrow x-\dfrac{1}{2}=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{1}{4}+\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{4}+\dfrac{2}{4}\)
\(\Rightarrow x=\dfrac{3}{4}\)
\(d,4^{x-3}+1=17\)
\(\Rightarrow4^{x-3}=17-1\)
\(\Rightarrow4^{x-3}=16\)
\(\Rightarrow4^{x-3}=4^2\)
\(\Rightarrow x-3=2\)
\(\Rightarrow x=2+3\)
\(\Rightarrow x=5\)
#Toru
`3/2 x -1 =1/2x -3/5`
`=> 3/2x -1/2x = -3/5 +1`
`=> 2/2x= -3/5 + 5/5`
`=> x= 2/5`
__
`1/2x +1/2(x-2) = 3/4 -2x`
`=> 1/2x + 1/2x - 2/2 = 3/4 -2x`
`=> 1/2x +1/2x +2x = 3/4 + 1`
`=> 1/2x +1/2x + 4/2x = 3/4 +4/4`
`=> 6/2x = 7/4`
`=> x= 7/4 : 3`
`=>x=7/12`
__
`(x-1/2) -1/4=0`
`=> x-1/2=1/4`
`=> x=1/4 +1/2`
`=> x= 1/4 +2/4`
`=>x=3/4`
__
`4^(x-3) +1=17`
`=> 4^(x-3) =17-1`
`=> 4^(x-3)=16`
`=> 4^(x-3)=4^2`
`=> x-3=2`
`=>x=2+3`
`=>x=5`
\(a,\dfrac{-5}{x-3}< 0\Leftrightarrow x-3>0\left(-5< 0\right)\Leftrightarrow x>3\\ b,\dfrac{3-x}{x^2+1}\ge0\Leftrightarrow3-x\ge0\left(x^2+1>0\right)\Leftrightarrow x\le3\\ c,\dfrac{\left(x-1\right)^2}{x-2}< 0\Leftrightarrow x-2< 0\left[\left(x-1\right)^2\ge0\right]\Leftrightarrow x< 2\)
a, \(A=\left(x+2y\right)^2-x+2y\)
Thay x = 2 ; y = -1 ta được
\(A=\left(2-2\right)^2-2-2=-4\)
b, Ta có \(\left(x^2+4>0\right)\left(x-1\right)=0\Leftrightarrow x=1\)
Thay x = 1 vào B ta được \(B=3+8-1=10\)
c, Thay x = 1 ; y = -1 ta được
\(C=3,2.1.\left(-1\right)=-3,2\)
d, Ta có \(x=\left|3\right|=3;y=-1\)Thay vào D ta được
\(D=3.9-5\left(-1\right)+1=27+5+1=33\)
thay x=2,y=-1 vào biểu thức A ta có;
A=(2+2.(-1)^2-2+2.(-1)
A=(2+-2)^2-2+-2
A=0-2+-2
A=-4
b)
(x^2+4)(x-1)=0
suy ra x-1=0(x^2+4>0 với mọi x thuộc thuộc R)
(+)x-1=0
x =1
thay x=1 vào biểu thức B ta có;
B=3.1^2+8.1-1
B=3.1+8-1
B=3+8-1
B=10
c)thay x=1 và y=-1 vào biểu thức C ta có;
C=3,2.1^5.(-1)^3
C=3,2.1.(-1)
C=(-3,2)
d)giá trị tuyệt đối của 3=3 hoặc (-3)
TH1;thay x=3:y=-1 vào biểu thức d ta có;
D=3.3^2-5.(-1)+1
D=3.9-(-5)+1
D=27+5+1
D=33
a)\(\frac{1}{3}+\frac{1}{2}:x=\frac{1}{5}\)
\(\frac{1}{2}:x=\frac{1}{5}-\frac{1}{3}\)
\(\frac{1}{2}:x=-\frac{2}{15}\)
\(x=-\frac{15}{4}\)
b)\(\frac{1}{3}x+\frac{2}{5}\left(x+1\right)=0\)
\(\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
\(\frac{11}{15}x=-\frac{2}{5}\)
\(x=-\frac{6}{11}\)
c)\(\left(x+\frac{1}{2}\right)\left(x-\frac{3}{4}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\x-\frac{3}{4}=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{4}\end{cases}}\)