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\(\text{a) A = | -x + 8| - 21}\)
Vì | -x + 8| \(\le\) 0 ( với mọi x )
=> A = | -x + 8| - 21\(\ge\) -21
=> Amax = -21 khi | -x + 8| = 0 => -x + 8 = 0 => -x = -8 => x = 8
Vậy với Amin = -21 thì x = 8
b) \(B=\left|-x-17\right|+\left|y-36\right|+12\)
Vì \(\left\{\begin{matrix}\left|-x-17\right|\ge0\\\left|y-36\right|\ge0\end{matrix}\right.\)=> \(\left|-x-17\right|+\left|y-36\right|\ge0\)
=> \(B=\left|-x-17\right|+\left|y-36\right|+12\le12\)
=> Bmin = 12 khi \(\left|-x-17\right|+\left|y-36\right|=0\)
=> \(\left\{\begin{matrix}\left|-x-17\right|=0\\\left|y-36\right|=0\end{matrix}\right.\)=> \(\left\{\begin{matrix}-x-17=0\\y-36=0\end{matrix}\right.\)=> \(\left\{\begin{matrix}-x=17\\y=36\end{matrix}\right.\)=>\(\left\{\begin{matrix}x=-17\\y=36\end{matrix}\right.\)
Vậy Bmin = 12 khi \(\left\{\begin{matrix}x=-17\\y=36\end{matrix}\right.\)
c) \(C=-\left|2x-8\right|-35\)
Vì \(-\left|2x-8\right|\ge0\)
=> \(C=-\left|2x-8\right|-35\ge-35\)
=> Cmin = -35 khi \(-\left|2x-8\right|=0\)=> \(-2x-8=0\)=>\(-2x=8\)=> \(x=4\)
Vậy Cmin = -35 khi x = 4
d) \(D=3\left(3x-12\right)^2-37\)
Vì \(\left(3x-12\right)^2\ge0\)
=> \(3\left(3x-12\right)^2\ge0\)
=> \(D=3\left(3x-12\right)^2-37\ge-37\)
=> Dmin = -37 khi \(3\left(3x-12\right)^2=0\) => \(\left(3x-12\right)^2=0\)=> \(3x-12=0\)=> \(3x=12\)=>\(x=4\)
Vậy Dmin = -37 khi x = 4
a, A=|-x+8|-21
Vì |-x+8|>hoặc =0 với mọi x
suy ra |-x+8|-21>hoặc = -21
Dấu = xảy ra khi và chỉ khi |-x+8|=0
Khi và chỉ khi -x+8=0
Khi và chỉ khi-x=-8
khi và chỉ khi x =8
Vậy GTNN của A là -21 tại x=8
Mấy ý này bản chất ko khác nhau nhé, mình làm mẫu, bạn làm tương tự mấy ý kia nhé
a, \(\left|5x\right|=x+2\)
Với \(x\ge0\)thì \(5x=x+2\Leftrightarrow x=\dfrac{1}{2}\)
Với \(x< 0\)thì \(5x=-x-2\Leftrightarrow6x=-2\Leftrightarrow x=-\dfrac{1}{3}\)
b, \(\left|7x-3\right|-2x+6=0\Leftrightarrow\left|7x-3\right|=2x-6\)
Với \(x\ge\dfrac{3}{7}\)thì \(7x-3=2x-6\Leftrightarrow5x=-3\Leftrightarrow x=-\dfrac{3}{5}\)( ktm )
Với \(x< \dfrac{3}{7}\)thì \(7x-3=-2x+6\Leftrightarrow9x=9\Leftrightarrow x=1\)( ktm )
Vậy phương trình vô nghiệm
e)
\(\left(x+3\right)^3=\left(x+3\right)^5\)
\(\Rightarrow\)\(x+3=1;0\)
TH1: TH2
\(x+3=0\) \(x+3=1\)
\(x=-3\) \(x=-2\)
\(x\in\left\{-3;-2\right\}\)
a) (x-1)(5x+3)=(3x-8)(x-1)
= (x-1)(5x+3)-(3x-8)(x-1)=0
=(x-1)[(5x+3)-(3x-8)]=0
=(x-1)(5x+3-3x+8)=0
=(x-1)(2x+11)=0
\(\Leftrightarrow\) x-1=0 hoặc 2x+11=0
\(\Leftrightarrow\) x=1 hoặc x=\(\dfrac{-11}{2}\)
Vậy S={1;\(\dfrac{-11}{2}\)}
b) 3x(25x+15)-35(5x+3)=0
=3x.5(5x+3)-35(5x+3)=0
=15x(5x+3)-35(5x+3)=0
=(5x+3)(15x-35)=0
\(\Leftrightarrow\) 5x+3=0 hoặc 15x-35=0
\(\Leftrightarrow\) x=\(\dfrac{-3}{5}\) hoặc x=\(\dfrac{7}{3}\)
Vậy S={\(\dfrac{-3}{5};\dfrac{7}{3}\)}
c) (2-3x)(x+11)=(3x-2)(2-5x)
=(2-3x)(x+11)-(3x-2)(2-5x)=0
=(3x-2)[(x+11)-(2-5x)]=0
=(3x-2)(x+11-2+5x)=0
=(3x-2)(6x+9)=0
\(\Leftrightarrow\) 3x-2=0 hoặc 6x+9=0
\(\Leftrightarrow\) x=\(\dfrac{2}{3}\) hoặc x=\(\dfrac{-3}{2}\)
Vậy S={\(\dfrac{2}{3};\dfrac{-3}{2}\)}
d) (2x2+1)(4x-3)=(2x2+1)(x-12)
=(2x2+1)(4x-3)-(2x2+1)(x-12)=0
=(2x2+1)[(4x-3)-(x-12)=0
=(2x2+1)(4x-3-x+12)=0
=(2x2+1)(3x+9)=0
\(\Leftrightarrow\)2x2+1=0 hoặc 3x+9=0
\(\Leftrightarrow\)x=\(\dfrac{1}{2}\)hoặc x=\(\dfrac{-1}{2}\) hoặc x=-3
Vậy S={\(\dfrac{1}{2};\dfrac{-1}{2};-3\)}
e) (2x-1)2+(2-x)(2x-1)=0
=(2x-1)[(2x-1)+(2-x)=0
=(2x-1)(2x-1+2-x)=0
=(2x-1)(x+1)=0
\(\Leftrightarrow\) 2x-1=0 hoặc x+1=0
\(\Leftrightarrow\) x=\(\dfrac{-1}{2}\) hoặc x=-1
Vậy S={\(\dfrac{-1}{2}\);-1}
f)(x+2)(3-4x)=x2+4x+4
=(x+2)(3-4x)=(x+2)2
=(x+2)(3-4x)-(x+2)2=0
=(x+2)[(3-4x)-(x+2)]=0
=(x+2)(3-4x-x-2)=0
=(x+2)(-5x+1)=0
\(\Leftrightarrow\) x+2=0 hoặc -5x+1=0
\(\Leftrightarrow\) x=-2 hoặc x=\(\dfrac{1}{5}\)
Vậy S={-2;\(\dfrac{1}{5}\)}
C=|2x-3/5|+4/3>=4/3
Dấu = xảy ra khi x=3/10
D=|x-3|+|-x-2|>=|x-3-x-2|=5
Dấu = xảy ra khi -2<=x<=3
\(a,3-x=x+1,8\)
\(\Rightarrow-x-x=1,8-3\)
\(\Rightarrow-2x=-1,2\)
\(\Rightarrow x=0,6\)
\(b,2x-5=7x+35\)
\(\Rightarrow2x-7x=35+5\)
\(\Rightarrow-5x=40\)
\(\Rightarrow x=-8\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(\Rightarrow2x+20=3x-18\)
\(\Rightarrow2x-3x=-18-20\)
\(\Rightarrow-x=-38\)
\(\Rightarrow x=38\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(\Rightarrow8x-3+1=1+6x+x\)
\(\Rightarrow8x-3=7x\)
\(\Rightarrow8x-7x=3\)
\(\Rightarrow x=3\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)
\(\Rightarrow-2x=\dfrac{10}{9}\)
\(\Rightarrow x=-\dfrac{5}{9}\)
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{16}{3}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(\Rightarrow x-4=5-x\)
\(\Rightarrow x+x=5+4\)
\(\Rightarrow2x=9\)
\(\Rightarrow x=\dfrac{9}{2}\)
\(k,7x^2-11=6x^2-2\)
\(\Rightarrow7x^2-6x^2=-2+11\)
\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
\(m,5\left(x+3\cdot2^3\right)=10^2\)
\(\Rightarrow5\left(x+24\right)=100\)
\(\Rightarrow x+24=20\)
\(\Rightarrow x=-4\)
\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)
#\(Urushi\text{☕}\)
a, A = 3,5 + |x - 2017| - 9
= -5,5 + |x - 2017|
Ta có : |x - 2017| \(\ge0\Rightarrow-5,5+\left|x-2017\right|\ge-5,5\)
Dấu ''='' xảy ra <=> x - 2017 = 0 <=> x = 2017
Vậy GTNN của A = -5,5 <=> x = 2017
@Cô Bé Dễ Thương
\(A=\left|-x+8\right|-21\)
\(A=\left|-x+8\right|-21\ge-21\)
\(MinA=-21\Leftrightarrow-x+8=0\)\(\Leftrightarrow x=8\)
\(B=\left|-x-17\right|+\left|y-36\right|+12\)
\(B=\left|-x-17\right|+\left|y-36\right|+12\ge12\)
\(MinB=12\Leftrightarrow\hept{\begin{cases}-x-17=0\\y-36=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-17\\y=36\end{cases}}\)
\(C=-\left|2x+8\right|-35\)
\(C=-\left|2x+8\right|-35\le-35\)
\(MaxC=-35\Leftrightarrow2x+8=0\Leftrightarrow x=-4\)
Trl
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