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a) \(3^x=81\)
\(3^x=3^4\)
\(\Rightarrow x=4\)
b) \(2^x.16=128\)
\(2^x=128:16\)
\(2^x=8\)
\(2^x=2^3\)
\(\Rightarrow x=3\)
c) \(3^x:9=27\)
\(3^x=27.9\)
\(3^x=243\)
\(3^x=3^5\)
\(\Rightarrow x=5\)
d) \(x^4=x\)
\(\Rightarrow x=0\)hoac \(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
e) \(\left(2x+1\right)^3=27\)
\(\left(2x+1\right)^3=3^3\)
\(\Rightarrow2x+1=3\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
f) \(\left(x-2\right)^2=\left(x-2\right)^4\)
\(\left(x-2\right)^2-\left(x-2\right)^4=0\)
\(\left(x-2\right)^2-\left(x-2\right)^2.\left(x-2\right)^2=0\)
\(\left(x-2\right)^2\left[1-\left(x-2\right)^2\right]=0\)
\(\left(x-2\right)^2\left(1-x+2\right)\left(1+x-2\right)=0\)
\(\Rightarrow\left(x-2\right)^2=0\)hoac \(\orbr{\begin{cases}3-x=0\\x-1=0\end{cases}}\)
\(\Rightarrow x-2=0\)hoac \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
\(\Rightarrow x=2\)hoac \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
a) \(3^x=81\Leftrightarrow3^x=3^4\Rightarrow x=4\)
b)\(2^x\times16=128\Leftrightarrow2^x=8\Leftrightarrow2^x=2^3\Rightarrow x=3\)
c) \(3^x\div9=27\Leftrightarrow3^x\div3^2=3^3\Rightarrow x=5\)
d) \(x^4=x\Leftrightarrow x=1\)
e) \(\left(2x+1\right)^3=27\Leftrightarrow\left(2x+1\right)^3=3^3\Rightarrow2x+1=3 \)
\(\Rightarrow2x=3+1\Leftrightarrow2x=4\Rightarrow x=2\)
F)
Câu 1:
\(\frac{x+16}{35}=\frac{x}{7}\)
\(\frac{x+16}{35}=\frac{5x}{35}\)
\(x+16=5x\)
\(5x-x=16\)
\(4x=16\)
\(x=\frac{16}{4}\)
\(x=4\)
Câu 2:
\(-2x^2+40=-10\)
\(-2x^2=-10-40\)
\(-2x^2=-50\)
\(x^2=\frac{-50}{-2}\)
\(x^2=25\)
\(x^2=\left(\pm5\right)^2\)
\(x=\pm5\)
Vậy x = 5 hoặc x = - 5.
Chúc bạn học tốt
1/a) Ta có: \(A=x^4+\left(y-2\right)^2-8\ge-8\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x=0\\y-2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\y=2\end{cases}}\)
Vậy GTNN của A = -8 khi x=0, y=2.
b) Ta có: \(B=|x-3|+|x-7|\)
\(=|x-3|+|7-x|\ge|x-3+7-x|=4\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x\ge3\\x\le7\end{cases}}\Rightarrow3\le x\le7\)
Vậy GTNN của B = 4 khi \(3\le x\le7\)
2/ a) Ta có: \(xy+3x-7y=21\Rightarrow xy+3x-7y-21=0\)
\(\Rightarrow x\left(y+3\right)-7\left(y+3\right)=0\Rightarrow\left(x-7\right)\left(y+3\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=7\\y=-3\end{cases}}\)
b) Ta có: \(\frac{x+3}{y+5}=\frac{3}{5}\)và \(x+y=16\)
Áp dụng tính chất bằng nhau của dãy tỉ số, ta có:
\(\frac{x+3}{y+5}=\frac{3}{5}\Rightarrow\frac{x+3}{3}=\frac{y+5}{5}=\frac{x+y+8}{8}=\frac{16+8}{8}=\frac{24}{8}=3\)
\(\Rightarrow\hept{\begin{cases}\frac{x+3}{3}=3\Rightarrow x+3=9\Rightarrow x=6\\\frac{y+5}{5}=3\Rightarrow y+5=15\Rightarrow y=10\end{cases}}\)
Bài 3: đề không rõ.
Bài 1:\(a,A=x^4+\left(y-2\right)^2-8\)
Có \(x^4\ge0;\left(y-2\right)^2\ge0\)
\(\Rightarrow A\ge0+0-8=-8\)
Dấu "=" xảy ra khi \(MinA=-8\Leftrightarrow x=0;y=2\)
\(b,B=\left|x-3\right|+\left|x-7\right|\)
\(\Rightarrow B=\left|x-3\right|+\left|7-x\right|\)
\(\Rightarrow B\ge\left|x-3+7-x\right|\)
\(\Rightarrow B\ge\left|-10\right|=10\)
Dấu "=" xảy ra khi \(MinB=10\Leftrightarrow3\le x\le7\Rightarrow x\in\left(3;4;5;6;7\right)\)
\(8-12x+6x^2-x^3\)
\(=\left(2-x\right)^3\)
\(125x^3-75x^2+15x-1\)
\(=\left(5x-1\right)^3\)
\(x^2-xz-9y^2+3yz\)
\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y-z\right)\)
\(x^3-x^2-5x+125\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
\(x^3+2x^2-6x-27\)
\(=x^3+5x^2+9x-3x^2-15x-27\)
\(=x\left(x^2+5x+9\right)-3\left(x^2+5x+9\right)\)
\(=\left(x-3\right)\left(x^2+5x+9\right)\)
\(12x^3+4x^2-27x-9\)
\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x+1\right)\left(4x^2-9\right)\)
\(=\left(3x+1\right)\left(2x-3\right)\left(2x+3\right)\)
\(4x^4+4x^3-x^2-x\)
\(=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=x\left(x+1\right)\left(4x^2-1\right)\)
\(=x\left(x+1\right)\left(2x-1\right)\left(2x+1\right)\)
1 a x=4
b x=-4
c x=-7
d x=3
e x=10
g x=60
h x=36
i x=16
2a 1,2,3,4,5,6,7,8,9
b 1,2,3,4,5,6,7,8,9.........
c rỗng
3a 0
b 0
c10