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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 ) \(\left(x+1\right)^2-3\left(x+1\right)^2=-8\)
\(\Leftrightarrow\left(x+1\right)^2.\left(1-3\right)=-8\)
\(\Leftrightarrow-2\left(x+1\right)^2=-8\)
\(\Leftrightarrow\left(x+1\right)^2=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
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.
Bài 1: Phá dấu ngoặc rồi tính:
a. \(\left(a+b+c\right)-\left(a-b+c\right)\)
\(=a+b+c-a+b-c\)
\(=\left(a-a\right)+\left(b+b\right)+\left(c-c\right)\)
\(=2b\)
b. \(\left(4x+5y\right)-\left(5x-4y-1\right)\)
\(=4x+5y-5x+4y+1\)
\(=\left(4x-5x\right)+\left(5y+4y\right)+1\)
\(=-x+9y+1\)
a)\(3^5.5^7.45=3^5.5^7.3^2.5=3^7.5^8\)
b)\(2^8.4^5.9^9\)\(=2^8.2^{10}.9^9=2^{18}.9^9\)
c)\(\left(2^3.3^5.5^7\right)^{10}.12^{20}=2^{13}.3^{15}.5^{17}.12^{20}\)\(=2^{13}.3^{15}.5^{17}.2^{40}.3^{20}=2^{53}.3^{35}.5^{17}\)
d)\(\left(x^2y\right)^5.\left(x^2y^2\right)^7.\left(x.y^2\right)^6.x^3=x^{10}.y^5.x^{14}.y^{14}.x^6.y^{12}.x^3\)
\(=x^{33}.y^{31}\)
e)\(18^{20}.45^5.5^{25}.8^{10}=2^{20}.3^{40}.5^5.3^{10}.5^5.5^{25}.2^{30}\)
\(=2^{50}.3^{50}.5^{35}=6^{50}.5^{35}\)
f)\(2^7.3^8.4^9.9^8=2^7.3^8.2^{18}.3^{16}=2^{25}.3^{24}\)
Bài 1:
a: \(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2.15+3.75=\dfrac{8}{5}\)
=>x+4/15=8/5 hoặc x+4/15=-8/5
=>x=4/3 hoặc x=-28/15
b: \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{3}x=-\dfrac{1}{6}\\\dfrac{5}{3}x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{6}:\dfrac{5}{3}=\dfrac{-3}{30}=\dfrac{-1}{10}\\x=\dfrac{1}{10}\end{matrix}\right.\)
c: \(\Leftrightarrow\left|x-1\right|-1=1\)
=>|x-1|=2
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1
Bài 2:
b: \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{9}{25}\)
Bài 3:
a: \(A=\left|x+\dfrac{15}{19}\right|-1>=-1\)
Dấu '=' xảy ra khi x=-15/19
b: \(\left|x-\dfrac{4}{7}\right|+\dfrac{1}{2}>=\dfrac{1}{2}\)
Dấu '=' xảy ra khi x=4/7
1.
A=\(\dfrac{3\left|x\right|+2}{\left|x\right|-5}=\dfrac{3\left|x\right|-15+17}{\left|x\right|-5}=\dfrac{3\left(\left|x\right|-5\right)+17}{\left|x\right|-5}=\dfrac{3\left(\left|x\right|-5\right)}{\left|x\right|-5}+\dfrac{17}{\left|x-5\right|}=3+\dfrac{17}{\left|x\right|-5}\)
Để A \(\in\)Z thì \(\left|x\right|-5\inƯ\left(17\right)=\left\{-17;-1;1;17\right\}\)
Ta có :
\(\left|x\right|-5=-17\Rightarrow\left|x\right|=-12\left(KTM\right)\)
\(\left|x\right|-5=-1\Rightarrow\left|x\right|=4\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
\(\left|x\right|-5=1\Rightarrow\left|x\right|=6\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
\(\left|x\right|-5=17\Rightarrow\left|x\right|=32\Rightarrow\left[{}\begin{matrix}x=32\\x=-32\end{matrix}\right.\)
Vậy để A \(\in\)Z thì x \(\in\) {-32;-6;-4;4;6;32}
bài 1
\(A=\left(\dfrac{3}{8}+\dfrac{1}{4}+\dfrac{5}{12}\right):\dfrac{7}{8}\)
\(A=\dfrac{9+6+10}{24}:\dfrac{7}{8}=\dfrac{25}{24}.\dfrac{8}{7}=\dfrac{25.1}{3.7}=\dfrac{25}{21}\)
\(B=\dfrac{1}{4}:\left(10,3-9,8\right)-\dfrac{3}{4}\)
\(B=\dfrac{1}{4}:\dfrac{1}{2}-\dfrac{3}{4}\)
\(B=\dfrac{1}{4}.2-\dfrac{3}{4}\)
\(B=\dfrac{1}{2}-\dfrac{3}{4}=-\dfrac{1}{4}\)
\(M=-\dfrac{5}{7}.\dfrac{2}{11}+\dfrac{5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\)
\(M=-\dfrac{5}{7}\left(-\dfrac{2}{11}+\dfrac{9}{11}\right)+1\dfrac{5}{7}\)
\(M=-\dfrac{5}{7}.\dfrac{7}{11}+\dfrac{12}{7}\)
\(M=-\dfrac{5}{11}+\dfrac{12}{7}=\dfrac{97}{77}\)
\(N=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2\)
\(N=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3.4}{16}\)
\(N=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{6}{7}-\dfrac{5}{8}=\dfrac{13}{56}\)
Bài 1:
a: \(\Leftrightarrow\dfrac{2}{3}\cdot\dfrac{6+9-4}{12}< =\dfrac{x}{18}< =\dfrac{7}{13}\cdot\dfrac{3-1}{6}\)
\(\Leftrightarrow\dfrac{22}{36}< =\dfrac{x}{18}< =\dfrac{14}{78}=\dfrac{7}{39}\)
\(\Leftrightarrow\dfrac{11}{9}< =\dfrac{x}{9}< =\dfrac{7}{13}\)
=>143<=x<=63
hay \(x\in\varnothing\)
b: \(\Leftrightarrow\dfrac{31\cdot9-26\cdot4}{180}\cdot\dfrac{-36}{35}< x< \dfrac{153+64+56}{168}\cdot\dfrac{8}{13}\)
\(\Leftrightarrow-1< x< 1\)
=>x=0
a.=\(\dfrac{4^3.9^3.5^44^4.18^2}{4^5.9^5.5^5}\)=\(\dfrac{4^4.9^2.2^2}{4^2.9^2.5}\)=\(\dfrac{4^2.2^2}{5}\)=\(\dfrac{64}{5}\)
Bài 2:
a) (2x+1)3 = 27
(2x+1)3 = 33
=> 2x+1 = 3
=> 2x = 2
=> x = 1