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\(\left(2x+1\right)\left(2x+3\right)\left(x+1\right)^2-18=0\)
\(\Leftrightarrow\left(x^2+2x+1\right)\left(4x^2+8x+3\right)-18=0\)
Đặt \(x^2+2x+1=a\ge0\)
\(\Rightarrow a\left(4a-1\right)-18=0\)
\(\Leftrightarrow4a^2-a-18=0\)
\(\Leftrightarrow\left(4a^2+8a\right)+\left(-9a-18\right)=0\)
\(\Leftrightarrow\left(a+2\right)\left(4a-9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=-2\left(l\right)\\a=\frac{9}{4}\end{cases}}\)
\(\Rightarrow x^2+2x+1=\frac{9}{4}\)
\(\Leftrightarrow4x^2+8x-5=0\)
\(\Leftrightarrow\left(4x^2-2x\right)+\left(10x-5\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{5}{2}\end{cases}}\)
a, ĐKXĐ: \(x\ne0;x\ne\pm1\)
\(P=\left(\frac{2x}{x^2-1}+\frac{x-1}{2x+2}\right):\frac{x+1}{2x}=\left(\frac{2x}{\left(x-1\right)\left(x+1\right)}+\frac{x-1}{2\left(x+1\right)}\right):\frac{x+1}{2x}\)
\(=\left(\frac{2x.2}{2\left(x-1\right)\left(x+1\right)}+\frac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}\right):\frac{x+1}{2x}\)
\(=\frac{4x+x^2-2x+1}{2\left(x-1\right)\left(x+1\right)}:\frac{x+1}{2x}=\frac{x^2+2x+1}{2\left(x-1\right)\left(x+1\right)}=\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}\cdot\frac{2x}{x+1}=\frac{x}{x-1}\)
b,Để \(P=2\Leftrightarrow\frac{x}{x-1}=2\Leftrightarrow2\left(x-1\right)=x\Leftrightarrow2x-2-x=0\Leftrightarrow x-2=0\Leftrightarrow x=2\left(tmđk\right)\)
Vậy để P=2 <=> x=2
a,\(A=x^2-x-1\)
\(=x^2-x+\frac{1}{4}-\frac{5}{4}\)
\(=\left(x-\frac{1}{2}\right)^2-\frac{5}{4}\)
Vì:\(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\forall x\)
Hay:\(A\ge0\forall x\)
Dấu = xảy ra khi:\(\left(x-\frac{1}{2}\right)^2=0\Rightarrow x=\frac{1}{2}\)
Vậy Min A=-5/4 tại x=1/2
Hai phần cn lại lm tg tự nha bn
\(a,x^2-4x+1=0.\)
\(\text{Áp dụng biệt thức }\Delta=b^2-4ac\text{, ta có:}\)(Lớp 9 kì 2 hok)
\(\Delta=-4^2-4.1.1=16-4=12\)
\(\Rightarrow\text{pt có 2 nghiệm }\orbr{\begin{cases}x_1=\frac{4-\sqrt{12}}{2}=2-\sqrt{3}\\x_2=\frac{4+\sqrt{12}}{2}=2+\sqrt{3}\end{cases}}\)
b,bn xem lại đề nếu đúng nói mk 1 tiếng mk làm tiếp cho
a) x3 + 2x - 3
=x3+x2+3x-x2+x+3
=x(x2+x+3)-1(x2+x+3)
=(x-1)(x2+x+3)
b) x3 - x2 + x + 3
=x3-2x2+3x+x2-2x+3
=x(x2-2x+3)+1(x2-2x+3)
=(x+1)(x2-2x+3)
c) 3x3 - 4x2 + 13x - 4
=3x3-3x2+12-x2-x+4
=3x(x2-x+4)-1(x2-x+4)
=(3x-1)(x2-x+4)
d) 6x3 + x2 + x + 1
=6x3-2x2+2x+3x2-x+1
=2x(3x2-x+1)+1(3x2-x+1)
=(2x+1)(3x2-x+1)
e)bạn phân tích tương tự nhé mk cho đáp án để bạn đổi chiếu nè
=(2x+1)(2x2+2x+1)
ĐKXĐ: \(x\ne1;x\ne-1\)
\(\dfrac{x+1}{2x-2}+\dfrac{-2x}{x^2-1}\) \(=\dfrac{x+1}{2\left(x-1\right)}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\) \(=\dfrac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}-\dfrac{4x}{2\left(x-1\right)\left(x+1\right)}\) \(=\dfrac{x^2+2x+1-4x}{2\left(x-1\right)\left(x+1\right)}\) \(=\dfrac{x^2-2x+1}{2\left(x-1\right)\left(x+1\right)}\) \(=\dfrac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}\) \(=\dfrac{x-1}{2\left(x+1\right)}\)