GIÚP MK VS
TÌM X
2^x+1-2^x=32
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a: \(49-y^2=6\left(x-2021\right)^2\)
=>\(49-y^2\ge0\) và \(49-y^2\) ⋮6
=>\(y^2\in\left\lbrace1;16;25;49\right\rbrace\)
TH1: \(y^2=1\)
Ta có: \(49-y^2=6\left(x-2021\right)^2\)
=>\(6\left(x-2021\right)^2=49-1=48\)
=>\(\left(x-2021\right)^2=8\)
mà x nguyên
nên x∈∅
TH2: \(y^2=16\)
Ta có: \(49-y^2=6\left(x-2021\right)^2\)
=>\(6\left(x-2021\right)^2=49-16=33\)
=>\(\left(x-2021\right)^2=5,5\)
mà x nguyên
nên x∈∅
TH3: \(y^2=25\)
Ta có: \(49-y^2=6\left(x-2021\right)^2\)
=>\(6\left(x-2021\right)^2=49-25=24\)
=>\(\left(x-2021\right)^2=4\)
=>x-2021=2 hoặc x-2021=-2
=>x=2023(nhận) hoặc x=2019(nhận)
\(y^2=25\)
=>y=5(nhận) hoặc y=-5(nhận)
TH4: \(y^2=49\)
Ta có: \(49-y^2=6\left(x-2021\right)^2\)
=>\(6\left(x-2021\right)^2=49-49=0\)
=>\(\left(x-2021\right)^2=0\)
=>x-2021=0
=>x=2021(nhận)
\(y^2=49\)
=>y=7(nhận) hoặc y=-7(nhận)
\(2^x.2^{x+1}=32\)
\(\Rightarrow2^{2x+1}=2^5\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=4\)\(\Rightarrow x=2\)
Vậy \(x=2\)
\(\frac{1}{x+1}+\frac{1}{x-1}+\frac{2}{1-x^2}\)
\(=\frac{1}{x+1}+\frac{1}{x-1}-\frac{2}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x-1+x+1-2}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{2x-2}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{2}{x+1}\)
e) Ta có: \(E=\left(3x+2\right)\left(3x-5\right)\left(x-1\right)\left(9x+10\right)+24x^2\)
\(=\left(9x^2-15x+6x-10\right)\left(9x^2+10x-9x-10\right)+24x^2\)
\(=\left(9x^2-10-9x\right)\left(9x^2-10+x\right)+24x^2\)
\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)-9x^2+24x^2\)
\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)+15x^2\)
\(=\left(9x^2-10\right)^2-3x\left(9x^2-10\right)-5x\left(9x^2-10\right)+15x^2\)
\(=\left(9x^2-10\right)\left(9x^2-3x-10\right)-5x\left(9x^2-10-3x\right)\)
\(=\left(9x^2-3x-10\right)\left(9x^2-5x-10\right)\)
2x+1 - 2x = 32
2x . 2-2x = 25
2x = 25
=> x =5
Vậy x =5
2^x + 1 - 2^x = 32
2^x . 2 - 2^x = 2^5
2^x + 2^x - 2^x = 2^5
2^x = 2^5
=> x = 5