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\(\left(x-2\right)\left(x^2+2x+4\right)+3x-4=\left(x+2\right)\left(x^2-2x+4\right)-x+1\)
\(\Rightarrow\left(x^3-8\right)+3x-4=\left(x^3+8\right)-x+1\)
\(\Rightarrow x^3-8+3x-4=x^3+8-x+1\)
\(\Rightarrow x^3-x^3+3x+x=8+8+4+1\)
\(\Rightarrow4x=21\)
\(\Rightarrow x=\dfrac{21}{5}\)
\(\sqrt{2x-1}=x^3-2x^2+2x\left(ĐK:x\ge\frac{1}{2}\right)\) \(\left(1\right)\)
\(\Leftrightarrow\sqrt{2x-1}=x^3-x\left(2x-1\right)+x\)
Đặt: \(\sqrt{2x-1}=a\left(a\ge0\right)\)
Khi đó pt (1) trở thành:
\(a=x^3-a^2x+x\)
\(\Leftrightarrow\left(x^3-a^2x\right)+\left(x-a\right)=0\)
\(\Leftrightarrow x\left(x-a\right)\left(x+a\right)+\left(x-a\right)=0\)
\(\Leftrightarrow\left(x-a\right)\left(x^2+ax+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-a=0\left(2\right)\\x^2+ax+1=0\left(3\right)\end{array}\right.\)
Giải (2): \(x-a=0\Leftrightarrow x=a\)
\(\Leftrightarrow x=\sqrt{2x-1}\)
\(\Leftrightarrow x^2=2x-1\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\left(tm\right)\)
Giải (3) \(x^2+ax+1=0\)
Vì: \(VT\left(3\right)>0\) ( Vì: \(x\ge\frac{1}{2};a\ge0\) )
\(VP\left(3\right)=0\)
=> pt(3) vô nghiệm
Vậy pt trình đã cho có tập nghiêm là \(S=\left\{1\right\}\)
BÀi này bn còn có thế lm bằng pp đưa chúng về tổng các bình phương bằng 0
\(1,x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\\ 2,-2x^2-x-1=-2\left(x^2+2\cdot\dfrac{1}{4}x+\dfrac{1}{16}+\dfrac{7}{16}\right)\\ =-2\left(x+\dfrac{1}{4}\right)^2-\dfrac{7}{8}\le-\dfrac{7}{8}< 0\\ 3,\dfrac{1}{2}x^2-2x+2=\dfrac{1}{2}\left(x^2-4x+4\right)=\dfrac{1}{2}\left(x-2\right)^2\ge0\)
\(7^{2x}+7^{2x+2}=2450\)
\(7^{2x}.1+7^{2x}.7^2=2450\)
\(7^{2x}.\left(1+7^2\right)=2450\)
\(7^{2x}.\left(1+49\right)=2450\)
\(7^{2x}.50=2450\)
\(7^{2x}=2450:50\)
\(7^{2x}=49\)
\(7^{2x}=7^2\)
\(\Rightarrow2x=2\)
\(x=2:2\)
\(x=1\)
Vậy \(x=1\)
\(7^{2x}+7^{2x+2}=2450\)
\(\Leftrightarrow7^{2x}+7^{2x}.7^2=2450\)
\(\Leftrightarrow7^{2x}.\left(1+7^2\right)=2450\)
\(\Leftrightarrow7^{2x}.50=2450\)
\(\Leftrightarrow7^{2x}=2450:50\Leftrightarrow7^{2x}=49\)
\(\Leftrightarrow7^{2x}=7^2\Leftrightarrow2x=2\)
\(\Leftrightarrow x=1\)
\(2x+7⋮2x-2\)
=>\(2x-2+9⋮2x-2\)
=>\(9⋮2x-2\)
=>\(2x-2\in\left\{1;-1;3;-3;9;-9\right\}\)
=>\(2x\in\left\{3;1;5;-1;11;-7\right\}\)
=>\(x\in\left\{\dfrac{3}{2};\dfrac{1}{2};\dfrac{5}{2};-\dfrac{1}{2};\dfrac{11}{2};-\dfrac{7}{2}\right\}\)
mà x nguyên
nên \(x\in\varnothing\)