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\(2,\dfrac{x-2}{2}-\dfrac{2}{3}\ge x-1\)
\(\Leftrightarrow\dfrac{x-2}{2}-\dfrac{2}{3}-\left(x-1\right)\ge0\)
\(\Leftrightarrow\dfrac{3\left(x-2\right)-4-6\left(x-1\right)}{6}\ge0\)
\(\Leftrightarrow3x-6-4-6x+6\ge0\)
\(\Leftrightarrow-3x-4\ge0\)
\(\Leftrightarrow-3x\ge4\)
\(\Leftrightarrow x\le-\dfrac{4}{3}\)
1a. th1: ta có x+2=3x+4 nếu x+2>0 => x > -2
x+2=3x+4
x-3x=4-2
-2x=2
x= -1 (chọn)
th2: ta có x+2= -3x+4 nếu x+2<0 => x< -2
x+2 = -3x+4
x+3x= 4-2
4x=2
x=\(\dfrac{1}{2}\)(loại)
S={-1}
1b. <=> \(\dfrac{2\left(2x+1\right)}{6}\) + \(\dfrac{3\left(x+1\right)}{6}\)= \(\dfrac{12}{6}\)
<=> 4x + 2 + 3x + 3= 12
<=> 7x= 7
<=> x= 1
S={1}
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B. 3.A. 4.C 5.C. 6.C
7.C. 8.C. 9.A. 10.D
11.D. 12.D. 13.B. 14.B
15.C. 16.A. 17.D
Mình ko bt đúng hay sai nên mong bạn thông cảm
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1
a
\(A=x^2-2x+4=x^2-2x+1+3\\=\left(x-1\right)^2+3\ge3\)
Min A = 3 khi và chỉ khi `x=1`
b
\(B=x^2+4x+5=x^2+4x+4+1\\ =\left(x+2\right)^2+1\ge1\)
Min B = 1 khi và chỉ khi `x=-2`
c
\(C=x^2+3x+4=x^2+3x+\dfrac{9}{4}+\dfrac{7}{4}\\ =\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)
Min C = \(\dfrac{7}{4}\) khi và chỉ khi \(x=-\dfrac{3}{2}\)
d
\(D=3x^2-6x+4=3\left(x^2-2x+\dfrac{4}{3}\right)\\ =3\left(x^2-2x+1+\dfrac{1}{3}\right)\\ =3\left(x-1\right)^2+3.\dfrac{1}{3}\ge3.\dfrac{1}{3}\)
Min D = 1 khi và chỉ khi `x=1`
Giúp mik vơi,HELPPPPP
Bài 6
a) x + 0,5 = 2/3
x + 1/2 = 2/3
x = 2/3 - 1/2
x = 1/6
b) 1/5 + (x - 2/3) = 5/3
x - 2/3 = 5/3 - 1/5
x - 2/3 = 22/15
x = 22/15 + 2/3
x = 32/15
c) (5/6 x + 3)² = 25/36
5/6 x + 3 = 5/6 hoặc 5/6 x + 3 = -5/6
*) 5/6 x + 3 = 5/6
5/6x = 5/6 - 3
5/6 x = -13/6
x = -13/6 : 5/6
x = -13/5
*) 5/6 x + 3 = -5/6
5/6 x = -5/6 - 3
5/6 x = -23/6
x = -23/6 : 5/6
x = -23/5
Vậy x = -23/5; x = -13/5
e) 2.|x - 1/8| = 6
|x - 1/8| = 6 : 2
|x - 1/8| = 3
*) Với x ≥ 1/8, ta có:
x - 1/8 = 3
x = 3 + 1/8
x = 25/8 (nhận)
*) Với x < 1/8, ta có:
x - 1/8 = -3
x = -3 + 1/8
x = -23/8 (nhận)
Vậy x = -23/8; x = 25/8
5:
2: \(\left|3x-5\right|-\dfrac{1}{7}=\dfrac{1}{3}\)
=>\(\left|3x-5\right|=\dfrac{1}{3}+\dfrac{1}{7}=\dfrac{10}{21}\)
=>\(\left[{}\begin{matrix}3x-5=\dfrac{10}{21}\\3x-5=-\dfrac{10}{21}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{115}{21}\\3x=-\dfrac{10}{21}+\dfrac{105}{21}=\dfrac{95}{21}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{115}{63}\\x=\dfrac{95}{63}\end{matrix}\right.\)
3:
\(\left(\dfrac{3}{5}x-\dfrac{2}{3}x-x\right)\cdot\dfrac{1}{7}=\dfrac{-5}{21}\)
=>\(x\left(\dfrac{3}{5}-\dfrac{2}{3}-1\right)=\dfrac{-5}{21}:\dfrac{1}{7}=\dfrac{-5}{21}\cdot7=-\dfrac{5}{3}\)
=>\(x\cdot\dfrac{9-10-15}{15}=\dfrac{-5}{3}\)
=>\(x\cdot\dfrac{-16}{15}=\dfrac{-5}{3}\)
=>\(x=\dfrac{5}{3}:\dfrac{16}{15}=\dfrac{5}{3}\cdot\dfrac{15}{16}=\dfrac{75}{48}=\dfrac{25}{16}\)
5:
\(0,2+\left|x-2,3\right|=1,1\)
=>\(\left|x-2,3\right|=1,1-0,2=0,9\)
=>\(\left[{}\begin{matrix}x-2,3=0,9\\x-2,3=-0,9\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=2,3+0,9=3,2\\x=2,3-0,9=1,4\end{matrix}\right.\)
6: \(5\left(x+2\right)^3+7=2\)
=>\(5\left(x+2\right)^3=-5\)
=>\(\left(x+2\right)^3=-1\)
=>x+2=-1
=>x=-3
8: \(14-\left|\dfrac{3}{2}x-1\right|=9\)
=>\(\left|\dfrac{3}{2}x-1\right|=14-9=5\)
=>\(\left[{}\begin{matrix}\dfrac{3}{2}x-1=5\\\dfrac{3}{2}x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{2}x=6\\\dfrac{3}{2}x=-4\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=6:\dfrac{3}{2}=6\cdot\dfrac{2}{3}=4\\x=-4:\dfrac{3}{2}=-4\cdot\dfrac{2}{3}=-\dfrac{8}{3}\end{matrix}\right.\)