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18 tháng 6 2022
a: \(\left(x^2+x\right)^2+2\left(x^2+x\right)-8=0\)
\(\Leftrightarrow\left(x^2+x+4\right)\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
hay \(x\in\left\{-2;1\right\}\)
b: \(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x+2\right)\left(x+4\right)+24=0\)
\(\Leftrightarrow\left(x^2+x-2\right)\left(x^2+x-12\right)+24=0\)
\(\Leftrightarrow\left(x^2+x\right)^2-14\left(x^2+x\right)+48=0\)
\(\Leftrightarrow\left(x^2+x-6\right)\left(x^2+x-8\right)=0\)
hay \(x\in\left\{-3;2;\dfrac{-1+\sqrt{33}}{2};\dfrac{-1-\sqrt{33}}{2}\right\}\)
1 tháng 7 2019
a) \(x^2-3x+4\)
\(=x^2-2\cdot x\cdot\frac{3}{2}+\frac{9}{4}+\frac{7}{4}\)
\(=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}>0\forall x\)
b) \(x^2-5x+8\)
\(=x^2-2\cdot x\cdot\frac{5}{2}+\frac{25}{4}+\frac{7}{4}\)
\(=\left(x-\frac{5}{2}\right)^2+\frac{7}{4}>0\forall x\)
c) \(x^2+y^2+2x-4x-4y+5\)
\(=\left(x+y\right)^2-4\left(x+y\right)+4+1\)
\(=\left(x+y-2\right)^2+1>0\forall x\)
10 tháng 6 2018
a )
\(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-5x-12+x^2-7x+10=3x^2-17x+20\)
\(\Leftrightarrow3x^2-12x-2-3x^2+17x-20=0\)
\(\Leftrightarrow5x-22=0\)
\(\Leftrightarrow x=\dfrac{22}{5}\)
b )
\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow-5x^2-2x+16+4x^2-4x-8+2x^2-8=0\)
\(\Leftrightarrow x^2-6x=0\)
\(\Leftrightarrow x\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
3 tháng 7 2019
a) (x+2)(x+3)-(x-2)(x+5)=0
\(x^2+3x+2x+6-x^2-5x+2x+10=0\)
\(2x+16=0\)
\(2x=-16\)
\(x=-8\)
Vậy......
b) (8-5x)(x+2)+4(x-2)(x+1)+2(x-2)(x+2)=0
\(8x+16-5x^2-10x+4x^2+4x-8x-8+2x^2+4x-4x-8=0\)
\(-6x+x^2=0\)
\(x\left(-6+x\right)=0\)
=> x=0 hoặc -6+x=0 <=>x=6
Vậy \(x\in\left\{0;6\right\}\)
3 tháng 7 2019
a) \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+2\right)x+\left(x+2\right).3-\left(x+5\right)x+\left(x+5\right).2=0\)
\(\Leftrightarrow x^2+2x+3x+6-x^2+5x+2x+10=0\)
\(\Leftrightarrow12x+16=0\)
\(\Leftrightarrow12x=-16\)
\(\Leftrightarrow x=\frac{-4}{3}\)
Vậy...
3 tháng 7 2019
\(a,\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)
\(x^2+5x+6-x^2-3x+10=0\)
\(2x+16=0\)
\(2x=-16\)
\(x=-8\)
\(b,\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)
\(8x+16-5x^2-10x+4x^2-4x-8+2x^2-8=0\)
\(x^2-6x=0\)
\(x\left(x-6\right)=0\)
\(\orbr{\begin{cases}x=0\\x-6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}}\)
3 tháng 7 2019
\(a,\)\(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Rightarrow x^2+5x+6-x^2-3x+10=0\)
\(\Rightarrow2x=-16\Leftrightarrow x=-8\)
\(b,\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow8x+16-5x^2-10x+4\left(x^2-x+2\right)+2\left(x^2-4\right)=0\)
\(\Rightarrow8x+16x-5x^2-10x+4x^2-4x+8+2x^2-8=0\)
\(\Rightarrow x^2+10x=0\Rightarrow x\left(x+10\right)=0\Rightarrow x\in\left\{0;-10\right\}\)
22 tháng 5 2022
Bài 5:
a: \(8A=8+8^2+...+8^8\)
\(\Leftrightarrow7A=8^8-1\)
hay \(A=\dfrac{8^8-1}{7}\)
b: \(8B=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(\Leftrightarrow8B=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(\Leftrightarrow8B=3^{16}-1\)
hay \(B=\dfrac{3^{16}-1}{8}\)