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1. A = (-2)(-3) - 5.|-5| + 125.\(\left(-\dfrac{1}{5}\right)^2\)
= 6 - 25 + 125.\(\dfrac{1}{25}\)
= -19 + 5
= -14
@Shine Anna
a: =>-3/2+x-7=5-1/3x+4/15
=>4/3x=413/30
hay x=413/40
b: \(\Leftrightarrow5-\dfrac{3}{2}x=-\dfrac{22}{3}\cdot\dfrac{-11}{8}=\dfrac{121}{12}\)
=>3/2x=-61/12
hay x=-61/18
c: (3x+2)2+|3x+2y|=0
=>3x+2=0 và 3x=-2y
=>x=-2/3 và -2y=-2
=>(x,y)=(-2/3;1)
a) \(2\left(x-5\right)-3\left(x+7\right)=14\)
\(\Leftrightarrow2x-10-3x-21=14\)
\(\Leftrightarrow-x-31=14\)
\(\Leftrightarrow-x=45\Leftrightarrow x=-45\)
b) \(5\left(x-6\right)-2\left(x+3\right)=12\)
\(\Leftrightarrow5x-30-2x-6=12\)
\(\Leftrightarrow3x-36=12\)
\(\Leftrightarrow3x=48\Leftrightarrow x=16\)
c) \(3\left(x-4\right)-\left(8-x\right)=12\)
\(\Leftrightarrow3x-12-8+x=12\)
\(\Leftrightarrow4x-20=12\)
\(\Leftrightarrow4x=32\Leftrightarrow x=8\)
d) \(-7\left(3x-5\right)+2\left(7x-14\right)=28\)
\(\Leftrightarrow-21x+35+14x-28=28\)
\(\Leftrightarrow-7x+35=0\Leftrightarrow x=5\)
\(A=\left|-x+8\right|-21\)
\(A=\left|-x+8\right|-21\ge-21\)
\(MinA=-21\Leftrightarrow-x+8=0\)\(\Leftrightarrow x=8\)
\(B=\left|-x-17\right|+\left|y-36\right|+12\)
\(B=\left|-x-17\right|+\left|y-36\right|+12\ge12\)
\(MinB=12\Leftrightarrow\hept{\begin{cases}-x-17=0\\y-36=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-17\\y=36\end{cases}}\)
\(C=-\left|2x+8\right|-35\)
\(C=-\left|2x+8\right|-35\le-35\)
\(MaxC=-35\Leftrightarrow2x+8=0\Leftrightarrow x=-4\)
\(\text{a) A = | -x + 8| - 21}\)
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
a ) \(\left(x+1\right)^2-3\left(x+1\right)^2=-8\)
\(\Leftrightarrow\left(x+1\right)^2.\left(1-3\right)=-8\)
\(\Leftrightarrow-2\left(x+1\right)^2=-8\)
\(\Leftrightarrow\left(x+1\right)^2=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
Vậy .......
b ) \(x^2-7x=4-7\left(x-3\right)\)
\(\Leftrightarrow x^2-7x-4+7x-21=0\)
\(\Leftrightarrow x^2-25=0\)
\(\Leftrightarrow x^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
Vậy ........
c ) \(\left(2x+1\right)^2-3x+3=4-3\left(x+1\right)\)
\(\Leftrightarrow\left(2x+1\right)^2-3\left(x-1\right)+3\left(x-1\right)=4\)
\(\Leftrightarrow\left(2x+1\right)^2=4\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=2\\2x+1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Vậy......
b. x2 - 7x = 4 - 7(x-3)
=> x2 - 7x = 4 - 7x +21
=> x2 - 7x + 7x = 25
=> x2 = 25
=> \(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
c.
\(B=\left|3x-7\right|-\left|3x+2\right|+8\)
Áp dụng tính chất:
\(\left|x\right|-\left|y\right|\le\left|x-y\right|\)
\(\left|3x-7\right|-\left|3x+2\right|\le\left|3x-7-3x-2\right|\)
\(B\le9+8=17\)
Dấu "=" xảy ra khi:
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}3x-7\ge0\Rightarrow3x\ge7\Rightarrow x\ge\dfrac{7}{3}\\3x+2\ge0\Rightarrow3x\ge-2\Rightarrow x\ge\dfrac{-2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}3x-7< 0\Rightarrow3x< 7\Rightarrow x< \dfrac{7}{3}\\3x+2< 0\Rightarrow3x< -2\Rightarrow x< -\dfrac{2}{3}\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x\ge\dfrac{7}{3}\) hoặc \(x< -\dfrac{2}{3}\)
Toshiro Kiyoshi bạn nói là nếu cả 2 số đều lớn hơn thì chọn số lớn hơn. Vì \(\dfrac{7}{3}>\dfrac{-2}{3}\) nên mk chọn là \(\dfrac{7}{3}\)Nhưng nếu \(x=2\) thì sao ?
Số đó \(< \dfrac{7}{3}\)