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Bài 1 : Tính nhanh
a) 16.(38−2)−38(16−1)16.(38−2)−38(16−1)
b) (−41).(59+2)+59(41−2)(−41).(59+2)+59(41−2)
Bài 2 :
Tìm các số x ; y ; x biết rằng :
x + y = 2 ; y + z = 3 ; z + x = -5
Bài 3 : Tìm x ; y ∈∈ Z biết rằng :
( y + 1 ) . xy - 1 ) = 3
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a) \(\left(x+\frac{1}{4}\right)^2+\frac{11}{25}=\frac{18}{25}\)
\(\Rightarrow\left(x+\frac{1}{4}\right)^2=\frac{7}{25}\)
\(\Rightarrow\) Không có x
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Bài 1:
\(\text{a) }x.x^2.x^3.x^4.x^5.....x^{49}.x^{50}\)
\(=x^{1+2+3+4+5+...+49+50}\)
\(=x^{\frac{51.50}{2}}\)
\(=x^{1275}\)
\(\text{b) Ta có:}\)
\(4^{15}=\left(2^2\right)^{15}=2^{2.15}=2^{30}\)
\(8^{11}=\left(2^3\right)^{11}=2^{3.11}=2^{33}\)
\(\text{Vì }2^{30}< 2^{33}\text{ nên }4^{15}< 8^{11}\)
Bài 2: Tìm x
\(\left(x-1\right)^4:3^2=3^6\)
\(\Rightarrow\left(x-1\right)^4=3^6\times3^2\)
\(\Rightarrow\left(x-1\right)^4=3^8\)
\(\Rightarrow\left(x-1\right)^4=3^{2.4}\)
\(\Rightarrow\left(x-1\right)^4=\left(3^2\right)^4\)
\(\Rightarrow x-1=9\)
\(\Rightarrow x=10\)
Bài 3 và bài 4 mk làm sau
Bài 1 : a) \(x.x^2.x^3.x^4.....x^{49}.x^{50}=x^{1+2+3+...+49+50}\) (Dễ rồi tự tính)
b) \(\hept{\begin{cases}4^{15}=\left(2^2\right)^{15}=2^{30}\\8^{11}=\left(2^3\right)^{11}=2^{33}\end{cases}}\)Rồi tự so sánh đi
Bài 2 :
\(\left(x-1\right)^4\div3^2=3^6\Leftrightarrow\left(x-1\right)^4=3^8=\left(3^2\right)^4=9^4\Leftrightarrow x-1=9\Leftrightarrow x=10\)
Bài 3 :
\(\hept{\begin{cases}27^{15}=\left(3^3\right)^{15}=3^{45}\\81^{11}=\left(3^4\right)^{11}=3^{44}\end{cases}}\) nt
Tìm x biết:
5) {x\(^2\) - [16\(^2\) - ( 8\(^2\) - 9 . 7)\(^3\) - 17 . 15 ]\(^3\) - 5 . 3 ]\(^3\) = 1
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\(\left\{x^2-\left[16^2-\left(8^2-9.7\right)^3-17.15\right]^3-5.3\right\}^3=1\)
\(\left\{x^2-\left[256-\left(64-9.7\right)^3-17.15\right]^3-5.3\right\}^3=1\)
\(\left\{x^2-\left[256-1^3-255\right]^3-15\right\}^3=1\)
\(\left\{x^2-\left[255-255\right]^3-15\right\}^3=1\)
\(\left\{x^2-0^3-15\right\}^3=1\)
\(\left\{x^2-15\right\}^3=1\)
\(\left\{4^2-15\right\}^3=1\)
\(\Rightarrow x=4\)
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5: \(\Leftrightarrow\left\{x^2-\left[256-\left(64-63\right)^3-255\right]^3-15\right\}^3=1\)
\(\Leftrightarrow\left\{x^2-15\right\}^3=1\)
\(\Leftrightarrow x^2-15=1\)
=>x=4 hoặc x=-4
a) \(15+\left|x-2\right|=-41+11\)
\(15+\left|x-2\right|=-30\)
\(\left|x-2\right|=-45\)( vô lí )
b) \(\left|x+3\right|-8-16=-17\)
\(\left|x+3\right|=-17+8+16\)
\(\left|x+3\right|=7\)
\(\Rightarrow\orbr{\begin{cases}x+3=7\\x+3=-7\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=-10\end{cases}}}\)
a,\(15+\left|x-2\right|=-41+11\)
\(\Rightarrow\left|x-2\right|=-41+11-15\)
\(\Rightarrow\left|x-2\right|=-45\)
Vì \(\left|x\right|\ge0\Rightarrow\left|x-2\right|=-45\)(vô lí)
b,\(\left|x+3\right|-8-16=-17\)
\(\Rightarrow\left|x+3\right|=-17+16+8\)
\(\Rightarrow\left|x+3\right|=7\)
=> x+3= 7 hoặc x+3=-7
=> x=4 hoặc x=-10