Câu hỏi của A Thuw

Question

Bài1:Thực hiện phép tính:a,(2^4.3.5^2):{450:[450-(4.5^3-2^3.5^2)]}b,3^3.5^2-20{90-[164-2.(7^8:7^6+7^0)]}c,[(18^7:18^6-17).2022-1986].5.1^2022-13^2.2020^0Bài2:Tìm x:a,(2^x+1)^2+3.(2^2+1)=2^2.10b,3.(x-7...

Toru · Accepted Answer

Bài 1.$a,\left(2^4\cdot3\cdot5^2\right):\left\{450:\left[450-\left(4\cdot5^3-2^3\cdot5^2\right)\right]\right\}$$=\left(16\cdot3\cdot25\right):\left\{450:\left[450-
\left(4\cdot125-8\cdot25\right)\right]\right\}$$=\left(48\cdot25\right):\left\{450:\left[450-\left(500-200\right)\right]\right\}$$=1200:\left[450:\left(450-300\right)\right]$$=1200:\left(450:150\right)$$=1200:3$$=400$$---$$b,3^3\cdot5^2-20\left\{90-\left[164-2\cdot\left(7^8:7^6+7^0\right)\right]\right\}$$=27\cdot25-20\left\{90-\left[164-2\cdot\left(7^2+1\right)\right]\right\}$$=675-20\left\{90-\left[164-2\cdot\left(49+1\right)\right]\right\}$$=675-20\left[90-\left(164-2\cdot50\right)\right]$$=675-20\left[90-\left(164-100\right)\right]$$=675-20\left(90-64\right)$$=675-20\cdot26$$=675-520$$=155$$---$$c,\left[\left(18^7:18^6-17\right)\cdot2022-1986\right]\cdot5\cdot1^{2022}-13^2\cdot2020^0$$=\left[\left(18-17\right)\cdot2022-1986\right]\cdot5\cdot1-169\cdot1$$=\left(1\cdot2022-1986\right)\cdot5-169$$=\left(2022-1986\right)\cdot5-169$$=36\cdot5-169$$=180-169$$=11$Bài 2.$a) (2^x+1)^2+3\cdot(2^2+1)=2^2\cdot10\\Rightarrow (2^x+1)^2+3\cdot(4+1)=4\cdot10\\Rightarrow (2^x+1)^2+3\cdot5=40\\Rightarrow (2^x+1)^2+15=40\\Rightarrow (2^x+1)^2=40-15\\Rightarrow (2^x+1)^2=25\\Rightarrow (2^x+1)^2= (\pm 5)^2\\Rightarrow 
\left[\begin{array}{}
2^x+1=5\
2^x+1=-5
\end{array} \right.\
\Rightarrow \left[\begin{array}{}
2^x=4\
2^x=-6 (vô.lí)
\end{array} \right. \
\Rightarrow 2^x=2^2\\Rightarrow x=2$Vậy $x=2$.$---$$b)3\cdot(x-7)+2\cdot(x+5)=41\\Rightarrow 3\cdot x+3\cdot(-7)+2\cdot x+2\cdot5=41\\Rightarrow 3x-21+2x+10=41\\Rightarrow (3x+2x)+(-21+10)=41\\Rightarrow 5x-11=41\\Rightarrow 5x=41+11\\Rightarrow 5x=52\\Rightarrow x=\dfrac{52}{5}$Vậy $x=\dfrac{52}{5}$.$Toru$Câu trả lời: Bài 1.$a,\left(2^4\cdot3\cdot5^2\right):\left\{450:\left[450-\left(4\cdot5^3-2^3\cdot5^2\right)\right]\right\}$$=\left(16\cdot3\cdot25\right):\left\{450:\left[450-
\left(4\cdot125-8\cdot25\right)\right]\right\}$$=\left(48\cdot25\right):\left\{450:\left[450-\left(500-200\right)\right]\right\}$$=1200:\left[450:\left(450-300\right)\right]$$=1200:\left(450:150\right)$$=1200:3$$=400$$---$$b,3^3\cdot5^2-20\left\{90-\left[164-2\cdot\left(7^8:7^6+7^0\right)\right]\right\}$$=27\cdot25-20\left\{90-\left[164-2\cdot\left(7^2+1\right)\right]\right\}$$=675-20\left\{90-\left[164-2\cdot\left(49+1\right)\right]\right\}$$=675-20\left[90-\left(164-2\cdot50\right)\right]$$=675-20\left[90-\left(164-100\right)\right]$$=675-20\left(90-64\right)$$=675-20\cdot26$$=675-520$$=155$$---$$c,\left[\left(18^7:18^6-17\right)\cdot2022-1986\right]\cdot5\cdot1^{2022}-13^2\cdot2020^0$$=\left[\left(18-17\right)\cdot2022-1986\right]\cdot5\cdot1-169\cdot1$$=\left(1\cdot2022-1986\right)\cdot5-169$$=\left(2022-1986\right)\cdot5-169$$=36\cdot5-169$$=180-169$$=11$Bài 2.$a) (2^x+1)^2+3\cdot(2^2+1)=2^2\cdot10\\Rightarrow (2^x+1)^2+3\cdot(4+1)=4\cdot10\\Rightarrow (2^x+1)^2+3\cdot5=40\\Rightarrow (2^x+1)^2+15=40\\Rightarrow (2^x+1)^2=40-15\\Rightarrow (2^x+1)^2=25\\Rightarrow (2^x+1)^2= (\pm 5)^2\\Rightarrow 
\left[\begin{array}{}
2^x+1=5\
2^x+1=-5
\end{array} \right.\
\Rightarrow \left[\begin{array}{}
2^x=4\
2^x=-6 (vô.lí)
\end{array} \right. \
\Rightarrow 2^x=2^2\\Rightarrow x=2$Vậy $x=2$.$---$$b)3\cdot(x-7)+2\cdot(x+5)=41\\Rightarrow 3\cdot x+3\cdot(-7)+2\cdot x+2\cdot5=41\\Rightarrow 3x-21+2x+10=41\\Rightarrow (3x+2x)+(-21+10)=41\\Rightarrow 5x-11=41\\Rightarrow 5x=41+11\\Rightarrow 5x=52\\Rightarrow x=\dfrac{52}{5}$Vậy $x=\dfrac{52}{5}$.$Toru$

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