Tính hợp lý giá trị biểu thức
A=(-3/4+2/3).11/9+(-1/4+1/3):|-9/11|
Giải giúp mk vs ạ mk cảm ơn trc ạ
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a, Ta có: \(A=\left|x+2\right|+\left|9-x\right|\ge\left|X+2+9-x\right|=11\)
Dấu "=' xảy ra khi \(\left(x+2\right)\left(9-x\right)\ge0\Leftrightarrow-2\le x\le9\)
Vậy MinA = 11 khi -2 =< x =< 9
b, Vì \(\left(x-1\right)^2\ge0\Rightarrow-\left(x-1\right)^2\le0\Rightarrow B=\frac{3}{4}-\left(x-1\right)^2\le\frac{3}{4}\)
Dấu "=" xảy ra khi x = 1
Vậy MaxB = 3/4 khi x=1
Ta có :\(A=\left|x+2\right|+\left|9-x\right|\ge\left|x+2+9-x\right|=11\)
Vậy \(A_{min}=11\) khi \(2\le x\le9\)
\(A=\dfrac{5}{11}.\dfrac{5}{7}+\dfrac{5}{11}.\dfrac{2}{7}+\dfrac{6}{11}=\dfrac{5}{11}\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\dfrac{6}{11}=\dfrac{5}{11}.1+\dfrac{6}{11}=\dfrac{5}{11}+\dfrac{6}{11}=\dfrac{11}{11}=1\)
\(B=\dfrac{3}{13}.\dfrac{6}{11}+\dfrac{3}{13}.\dfrac{9}{11}-\dfrac{3}{13}.\dfrac{4}{11}=\dfrac{3}{13}\left(\dfrac{6}{11}+\dfrac{9}{11}-\dfrac{4}{11}\right)=\dfrac{3}{13}.1=\dfrac{3}{13}\)
\(C=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right)\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right).0=0\)
(x+2)(4x^2 -2x+1)+(3-2x)(9+6x+4x^2) = -4x^3+6x^2-3x+29
nha bạn chúc bạn học tốt nha
-3/11.(-22)/66.121/15
=(-3).(-22).121
11.66.15
=11
15
3/7.2/5.7/3.20.19/72
=3.2.7.20.19
7.5.3.72
=76
16
6/7.8/13+6/13.9/7-3/13.6/7
=6/7.8/13+6/7.9/13-3/13.6/7
=6/7.(8/13+9/13-3/13)
=6/7.14/13
=12/13
-1/4.152/11+68/4.(-1)/11
=152/4.(-1)/11+68/4.(-1)/11
=(-1)/11.(152/4+68/4)
=(-1)/11.220/4
=-110/22
-5/7.2/11+(-5)/7.9/11+12/7
=-5/7.2/11+-5/7.9/11+12/7
=-5/7.(2/11+9/11)+12/7
=-5/7.1+12/7
=(-5)/7+12/7
=7/7
=1
146/13-(18/7+68/13)
=146/13-18/7-68/13
=(146/13-68/13)-18/7
=78/13-18/7
=6-18/7
=42/7-18/7
=24/7
\(A=100+98+96+...+2-97-95-...-1\)
\(A=100+\left(98-98\right)+\left(96-95\right)+...+\left(2-1\right)\)
\(A=100+1+1+...+1\)
\(A=100+1\cdot49\)
\(A=100\cdot49\)
\(A=4900\)
\(B=1+2-3-4+5+6-7-8+9+10-11-12+...-299-300+301+302\)
\(B=1+\left(2-3-4+5\right)+\left(6-7-8+9\right)+...+\left(298-299-300+301\right)+302\)
\(B=1+0+0+...+302\)
\(B=1+302\)
\(B=303\)
b) Để \(B=\frac{x+2}{x+1}\)có giá trị nguyên thì \(x+2⋮x+1\)
Ta có : \(x+2⋮x+1\)
\(\Rightarrow x+1+1⋮x+1\)
Mà \(x+1⋮x+1\)
\(\Rightarrow1⋮x+1\)
\(\Rightarrow x+1\inƯ\left(1\right)=\left\{\pm1\right\}\)
+) x+1=-1\(\Rightarrow\)x=-2 (thỏa mãn)
+) x+1=1\(\Rightarrow\)x=0 (thỏa mãn)
Vậy \(x\in\left\{-2;0\right\}\)
Các phần sau bạn làm tương tự nhé!
Học tốt!
#Huyền#
\(A=\frac{15\times3^{11}+4\times27^4}{9^7}\)
\(A=\frac{15\times177147+4\times531441}{4782969}\)
\(A=\frac{2657205+2125764}{4782969}\)
\(A=\frac{47829969}{47829969}=1\)
A=(-3/4+2/3).11/9+(-1/4+1/3):|-9/11|
\(=\left(-\frac{3}{4}+\frac{2}{3}\right).\frac{11}{9}+\left(-\frac{1}{4}+\frac{1}{3}\right).\frac{11}{9}\)
\(=\frac{11}{9}\left(-\frac{3}{4}+\frac{2}{3}-\frac{1}{4}+\frac{1}{3}\right)\)
\(=\frac{11}{9}\left(-1+1\right)\)
\(=\frac{11}{9}.0\)
=0