g \(2\left(x+2\right)^2\ge2x+\frac{7}{2}\)
h \(\frac{x^2+x-1}{1-x}>-x\)
\(|x^2+x+12|>x^2+x+12\)
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Những câu hỏi liên quan
AH
Akai Haruma
Giáo viên
29 tháng 4 2019
a)
ĐKXĐ: \(x\neq 0; x\neq -10\)
\(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\)
\(\Leftrightarrow \frac{x+10+x}{x(x+10)}=\frac{1}{12}\)
\(\Leftrightarrow \frac{2x+10}{x(x+10)}=\frac{1}{12}\)
\(\Rightarrow 12(2x+10)=x(x+10)\)
\(\Leftrightarrow x^2-14x-120=0\)
\(\Leftrightarrow (x+6)(x-20)=0\Rightarrow \left[\begin{matrix} x=-6\\ x=20\end{matrix}\right.\) (đều thỏa mãn)
b)
ĐKXĐ: \(x\neq 0; x\neq 3\)
PT\(\Leftrightarrow \frac{(x+3).x-(x-3)}{x(x-3)}=\frac{3}{x(x-3)}\)
\(\Leftrightarrow \frac{x^2+2x+3}{x(x-3)}=\frac{3}{x(x-3)}\)
\(\Rightarrow x^2+2x+3=3\)
\(\Leftrightarrow x^2+2x=0\Leftrightarrow x(x+2)=0\Rightarrow \left[\begin{matrix} x=0\\ x=-2\end{matrix}\right.\) . Kết hợp với đkxđ suy ra $x=-2$
AH
Akai Haruma
Giáo viên
29 tháng 4 2019
c)
ĐKXĐ: \(x\neq \pm 2\)
\(\frac{3}{x+2}-\frac{2}{x-2}+\frac{8}{x^2-4}=0\)
\(\Leftrightarrow \frac{3(x-2)-2(x+2)}{(x+2)(x-2)}+\frac{8}{x^2-4}=0\)
\(\Leftrightarrow \frac{x-10}{x^2-4}+\frac{8}{x^2-4}=0\)
\(\Leftrightarrow \frac{x-2}{x^2-4}=0\Leftrightarrow \frac{1}{x+2}=0\) (vô lý)
Vậy pt vô nghiệm.
d)
ĐKXĐ: \(x\neq -2; x\neq 3\)
PT \(\Leftrightarrow \frac{3(x-3)-2(x+2)}{(x+2)(x-3)}=\frac{8}{(x-3)(x+2)}\)
\(\Leftrightarrow \frac{x-13}{(x+2)(x-3)}=\frac{8}{(x-3)(x+2)}\)
\(\Rightarrow x-13=8\Rightarrow x=21\) (thỏa mãn)
Vậy..........
12 tháng 3 2020
b) Ta có: \(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)
⇔\(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)-\left(x+1\right)^3=0\)
⇔\(x^3-6x^2+12x-8+9x^2-1-\left(x^3+3x^2+3x+1\right)=0\)
⇔\(x^3+3x^2+12x-9-x^3-3x^2-3x-1=0\)
⇔\(9x-10=0\)
hay 9x=10
⇔\(x=\frac{10}{9}\)
Vậy: \(x=\frac{10}{9}\)
c) \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{5}\)
⇔\(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{5}=0\)
⇔\(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}-\frac{3\left(x+7\right)}{15}=0\)
⇔\(3\left(2x-1\right)-5\left(x-2\right)-3\left(x+7\right)=0\)
⇔\(6x-3-5x+10-3x-21=0\)
⇔\(-2x-14=0\)
⇔\(-2x=14\)
hay x=-7
Vậy: x=-7
d) \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}=\frac{13x+4}{21}\)
⇔\(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)
⇔\(\frac{6\left(x-3\right)}{21}+\frac{7\left(x-5\right)}{21}-\frac{13x+4}{21}=0\)
⇔\(6x-18+7x-35-13x-4=0\)
⇔\(-21\ne0\)
Vậy: x∈∅
e) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
⇔\(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}-\frac{\left(x+10\right)\left(x-2\right)}{3}=0\)
⇔\(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{3\left(x+4\right)\left(2-x\right)}{12}-\frac{4\left(x+10\right)\left(x-2\right)}{12}=0\)
⇔\(x^2+14x+40-\left(3x+12\right)\left(2-x\right)-\left(4x+40\right)\left(x-2\right)=0\)
⇔\(x^2+14x+40-\left(24-6x-3x^2\right)-\left(4x^2+32x-80\right)=0\)
⇔\(x^2+14x+40-24+6x+3x^2-4x^2-32x+80=0\)
⇔\(-12x+96=0\)
⇔\(-12x=-96\)
hay x=8
Vậy: x=8
8 tháng 1 2020
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
a/ \(\Leftrightarrow2\left(x^2+4x+4\right)-2x-\frac{7}{2}\ge0\)
\(\Leftrightarrow2x^2+6x+\frac{9}{2}\ge0\)
\(\Leftrightarrow2\left(x+\frac{3}{2}\right)^2\ge0\) (luôn đúng)
Vậy nghiệm của BPT là \(D=R\)
b/ \(\Leftrightarrow\frac{x^2+x-1}{1-x}+x>0\)
\(\Leftrightarrow\frac{x^2+x-1+x-x^2}{1-x}>0\)
\(\Leftrightarrow\frac{2x-1}{1-x}>0\Rightarrow\frac{1}{2}< x< 1\)
c/ \(x^2+x+12=\left(x+\frac{1}{2}\right)^2+\frac{47}{4}>0\) \(\forall x\) nên BPT tương đương:
\(x^2+x+12>x^2+x+12\)
\(\Leftrightarrow0>0\)
Vậy BPT vô nghiệm