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a) \(2.5^2.3^2+\left\{\left[2.5^3-\left(5x+4\right).5\right]:\left(2^2.3.5\right)\right\}=453\)
\(2.25.9+\left\{\left[2.125-\left(5x+4\right).5\right]:\left(4.3.5\right)\right\}=453\)
\(50.9+\left\{\left[250-\left(5x+4\right).5\right]:60\right\}=453\)
\(450+\left\{\left[250-\left(5x+4\right).5\right]:60\right\}=453\)
\(\left[250-\left(5x+4\right).5\right]:60=453-450\)
\(\left[250-\left(5x+4\right).5\right]:60=3\)
\(250-\left(5x+4\right).5=3.60\)
\(250-\left(5x+4\right).5=180\)
\(\left(5x+4\right).5=250-180\)
\(\left(5x+4\right).5=70\)
\(5x+4=70:5\)
\(5x+4=14\)
\(5x=14-4\)
\(5x=10\)
\(x=10:5\)
\(x=2\)
Vậy \(x=2\)
b) \(\left|x-\frac{1}{3}\right|=2\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=-2\\x-\frac{1}{3}=2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\left(-2\right)+\frac{1}{3}\\x=2+\frac{1}{3}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-5}{3}\\x=\frac{7}{3}\end{cases}}\)
Vậy \(x\in\left\{\frac{-5}{7};\frac{7}{3}\right\}\)
\(5^{2x-3}-2.5^2=5^2.3\)
\(\Rightarrow5^{2x-3}=5^2.3+5^2.2\)
\(\Rightarrow5^{2x-3}=5^2.\left(2+3\right)=5^3\)
\(\Rightarrow2x-3=3\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
\(5^{2x-3}\)-2.\(5^2\)= \(5^2\).3
\(5^{2x-3}\) _2.25=25.3
\(5^{2x-3_{ }}\)_50=75
\(5^{2x-3}\)= 75_50
\(5^{2x-3}\)= \(5^2\)
suy ra : 2x-3=2
2x=2.3
2x=6
x=6:2
x=3
\(\left(19x+2.5^2\right):14=\left(13-8\right)^2-4^2\)
\(\Rightarrow\left(19x+50\right):14=25-16\)
\(\Rightarrow19x+50=9.14=126\)
\(\Rightarrow19x=76\)
\(\Rightarrow x=76:19=4\)
Ta có :
\(\frac{19x+2.5^2}{14}=\left(13-8\right)^2-4^2\)
\(\Leftrightarrow\)\(\frac{19x+2.25}{14}=5^2-4^2\)
\(\Leftrightarrow\)\(\frac{19x+50}{14}=25-16\)
\(\Leftrightarrow\)\(\frac{19x+50}{14}=9\)
\(\Leftrightarrow\)\(19x+50=126\)
\(\Leftrightarrow\)\(19x=76\)
\(\Leftrightarrow\)\(x=\frac{76}{19}\)
\(\Leftrightarrow\)\(x=4\)
Vậy \(x=4\)
c,\(43+x=2.5^2-\left(x-57\right)\)
\(< =>43+x=50-x+57\)
\(< =>2x=50+57-43\)
\(< =>x=\frac{107-43}{2}=32\)
d,\(-3.2^2\left(x-5\right)+7\left(3-x\right)=5\)
\(< =>-12.\left(x-5\right)+7.\left(3-x\right)=5\)
\(< =>-12x+60+21-7x=5\)
\(< =>-19x=5-81=-76\)
\(< =>x=-\frac{76}{-19}=4\)
Bài 2:
a) \(A=\left|x-3\right|+10\)
Vì \(\left|x-3\right|\ge0\forall x\)\(\Rightarrow\left|x-3\right|+10\ge10\forall x\)
hay \(A\ge10\)
Dấu " = " xảy ra \(\Leftrightarrow x-3=0\)\(\Leftrightarrow x=3\)
Vậy \(minA=10\Leftrightarrow x=3\)
b) \(B=-7+\left(x-1\right)^2\)
Vì \(\left(x-1\right)^2\ge0\forall x\)\(\Rightarrow-7+\left(x-1\right)^2\ge-7\forall x\)
hay \(B\ge-7\)
Dấu " = " xảy ra \(\Leftrightarrow x-1=0\)\(\Leftrightarrow x=1\)
Vậy \(minB=-7\Leftrightarrow x=1\)
Tìm x biết : 22.5-[7.(x-3)+22]=2
Bài Làm
22.5-[7.(x-3)+22]=2
=>4.5-[7.(x-3)+4]=2
=>20-[7.(x-3)+4]=2
=>7.(x-3)+4=20-2
=>7.(x-3)+4=18
=>7.(x-3)=18-4
=>7.(x-3)=14
=>x-3=14:7
=>x-3=2
=>x=2+3
=>x=5
Vậy x = 5