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câu c nè
\(\frac{x^2-3x+5}{x+1}=\frac{\left(x^2+2x+1\right)-5x+4}{x+1}=\frac{\left(x+1\right)^2-5\left(x+1\right)+9}{x+1}\)
Ta có \(\frac{x+2}{x+1}=\frac{\left(x+1\right)+1}{x+1}=1+\frac{1}{x+1}\)
\(A=\frac{x+6}{x-2}\)ĐKXĐ : \(x\ne2\)
\(=\frac{x-2+8}{x-2}=\frac{8}{x-2}\)
Suy ra : \(x-2\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
| x - 2 | 1 | -1 | 2 | -2 | 4 | -4 | 8 | -8 |
| x | 3 (tm) | 1 (tm) | 4 (tm) | 0 (tm) | 6 (tm) | -2 (tm) | 10 (tm) | -6 (tm) |
c. x+1/94 + x+2/93 = x+3/92 + x+4/91
\(a,\left(2x-1\right)^2-\left(2x+1\right)^2=4\left(x-3\right)\)
\(\Leftrightarrow4x^2-4x+1-4x^2-4x-1=4x-12\)
\(\Leftrightarrow-12x=-12\)
\(\Leftrightarrow x=1\)
\(b,\left(\frac{5x-7}{2}\right)=\left(\frac{16x+1}{7}\right)\)
\(\Leftrightarrow35x-49=32x+2\)
\(\Leftrightarrow3x=51\Leftrightarrow x=17\)
x^2+4x+5 ÷ -3-+9<0
Đáng lẽ dấu chia là dấu gạch ở dưới mà mình không biết cách làm nên mình ghi dấu ÷
\(\frac{x-1}{2}\cdot\frac{x+1}{2}\cdot(4x-1)\)
\(=\frac{\left(x-1\right)\left(x+1\right)\left(4x-1\right)}{2\cdot2}\)
\(=\frac{(x^2-1)\left(4x-1\right)}{4}\)
\(=\frac{4x^3-x^2-4x+1}{4}\)
Tìm GTNN của: \(\frac{x^2-3}{x^2+1}\)
Ta có: \(\frac{x^2-3}{x^2+1}=\frac{x^2+1-4}{x^2+1}=1-\frac{4}{x^2+1}\)
Có: \(x^2+1\ge1\)=> \(\frac{4}{x^2+1}\le\frac{4}{1}=4\) => \(1-\frac{4}{x^2+1}\ge1-4=-3\)
=> \(\frac{x^2-3}{x^2+1}\ge-3\)
Dấu "=" xảy ra <=> x ^2 + 1 = 1 <=> x^2 = 0 <=> x = 0
Vậy GTNN của \(\frac{x^2-3}{x^2+1}\)là -3 tại x = 0
a) \(\left(2x+1\right)^2-4\left(x+2\right)^2=12\)
\(\Leftrightarrow4x^2+4x+1-4\left(x^2+4x+4\right)=12\)
\(\Leftrightarrow4x^2+4x+1-4x^2-16x-16-12=0\)
\(\Leftrightarrow-12x-27=0\)
\(\Leftrightarrow x=\frac{-9}{4}\)
b) xem lại đề
c) \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x-3\right)\left(3-x\right)=1\)
\(\Leftrightarrow x^3-27-x\left(x-3\right)^2=1\)
\(\Leftrightarrow x^3-27-x\left(x^2-6x+9\right)-1=0\)
\(\Leftrightarrow x^3-28-x^3+6x^2-9x=0\)
\(\Leftrightarrow6x^2-9x-28=0\)
\(\Leftrightarrow6\left(x^2-\frac{3}{2}x-\frac{14}{3}\right)=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\frac{3}{4}+\frac{9}{16}-\frac{251}{48}=0\)
\(\Leftrightarrow\left(x-\frac{3}{4}\right)^2=\frac{251}{48}=\left(\pm\sqrt{\frac{251}{48}}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{3}{4}=\sqrt{\frac{251}{48}}=\frac{\sqrt{753}}{12}\\x-\frac{3}{4}=-\sqrt{\frac{251}{48}}=\frac{-\sqrt{753}}{12}\end{matrix}\right.\)
\(\Leftrightarrow x=\frac{\pm\sqrt{753}}{12}+\frac{3}{4}=\frac{9\pm\sqrt{753}}{12}\)
d) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-19\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+19=0\)
\(\Leftrightarrow12x+15=0\)
\(\Leftrightarrow x=\frac{-5}{4}\)
Theo giả thiết:
\(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=3ab+3bc+3ca\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Dễ thấy \(VT\ge0\forall a;b;c\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\)\(\Leftrightarrow a=b=c\)(đpcm)
ĐKXĐ: x≠-3; x≠3; x≠6
Ta có: \(\frac{x}{x+3}-\frac{x-2}{x-6}=\frac{x+2}{x^2-9}\)
\(\Leftrightarrow\frac{x\cdot\left(x-3\right)\cdot\left(x-6\right)}{\left(x+3\right)\left(x-3\right)\left(x-6\right)}-\frac{\left(x-2\right)\left(x+3\right)\left(x-3\right)}{\left(x-6\right)\left(x+3\right)\left(x-3\right)}-\frac{\left(x+2\right)\left(x-6\right)}{\left(x-3\right)\left(x+3\right)\left(x-6\right)}=0\)
\(\Leftrightarrow x^3-6x^2-3x^2+18x-\left(x^3-2x^2-9x+18\right)-\left(x^2-4x-12\right)=0\)
\(\Leftrightarrow x^3-6x^2-3x^2+18x-x^3+2x^2+9x-18-x^2+4x+12=0\)
\(\Leftrightarrow-8x^2+31x-6=0\)
Δ=\(31^2-4\cdot\left(-8\right)\cdot\left(-6\right)=769\)
Vì Δ>0
nên \(\left\{{}\begin{matrix}x_1=\frac{-31-\sqrt{769}}{2\cdot\left(-8\right)}=\frac{31+\sqrt{769}}{16}\\x_2=\frac{-31+\sqrt{769}}{2\cdot\left(-8\right)}=\frac{31-\sqrt{769}}{16}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{\frac{31+\sqrt{769}}{16};\frac{31-\sqrt{769}}{16}\right\}\)