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tôi bt làm 1 câu à mấy câu kia khó quá *-*
1. 5x2+4x-2=0
\(\Leftrightarrow x\left(5x+4\right)=2\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\5x+4=2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{-2}{5}\end{cases}}}\)
\(\Rightarrow\) Nghiệm pt là :\(S=\left\{\frac{-2}{5};2\right\}\)
chúc bn sớm làm dc bài này ha
\(2x^2-3x-2\)
\(=2x^2-4x+x-2\)
\(=2x\left(x-2\right)+\left(x-2\right)\)
\(=\left(x-2\right)\left(2x+1\right)\)
\(3x^2+x-2\)
\(=3x^2+3x-2x-2\)
\(=3x\left(x+1\right)-2\left(x-1\right)\)
\(=\left(x-1\right)\left(3x-2\right)\)
\(4x^2-7x-2\)
\(=4x^2-8x+x-2\)
\(=4x\left(x-2\right)+x-2\)
\(=\left(x-2\right)\left(4x+1\right)\)
\(4,4x^2+5x-6=4x^2+8x-3x-6\)
\(=4x\left(x+2\right)-3\left(x+2\right)=\left(4x-3\right)\left(x+2\right)\)
\(5,\) \(4x^2+15x+9=4x^2+12x+3x+9\)
\(=4x\left(x+3\right)+3\left(x+3\right)\)
\(=\left(4x+3\right)\left(x+3\right)\)
PTĐTTNT?
1.Đặt \(a^2+a=t\)
\(\Rightarrow\left(a^2+a\right)\left(a^2+a+1\right)-2\)
\(=t\left(t+1\right)-2\)
\(=t^2+t-2\)
\(=t^2+2t-\left(t+2\right)\)
\(=t\left(t+2\right)-\left(t+2\right)\)
\(=\left(t+2\right)\left(t-1\right)\)
Sửa đề:
\(x^4+2011x^2+2010x+2011\)
\(=\left(x^4-x\right)+2011x^2+2011x+2011\)
\(=x\left(x^3-1\right)+2011\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2011\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2011\right)\)
3. \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-120\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-120\)
Đặt \(x^2+5x+4=t\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-120\)
\(=t\left(t+2\right)-120\)
\(=t^2+2t+1-121\)
\(=\left(t+1\right)^2-11^2\)
\(=\left(t+1-11\right)\left(t+1+11\right)\)
\(=\left(t-10\right)\left(t+12\right)\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+16\right)\)
\(=\left[\left(x^2-x\right)+\left(6x-6\right)\right]\left(x^2+5x+16\right)\)
\(=\left[x.\left(x-1\right)+6\left(x-1\right)\right]\left(x^2+5x+16\right)\)
\(=\left(x-1\right)\left(x+6\right)\left(x^2+5x+16\right)\)
4. \(\left(x^2+x+4\right)^2+8x\left(x^2+x+1\right)+15x^2\)
\(=\left(x^2+x+4\right)^2+2.\left(x^2+x+1\right).4x+\left(4x\right)^2-x^2\)
\(=\left(x^2+x+4+4x\right)^2-x^2\)
\(=\left(x^2+4+5x-x\right)\left(x^2+5x+x+4\right)\)
\(=\left(x^2+4x+4\right)\left(x^2+6x+4\right)\)
\(=\left(x+2\right)^2\left[\left(x^2+2.x.3+3^2\right)-\left(\sqrt{5}\right)^2\right]\)
\(=\left(x+2\right)^2\left[\left(x+3\right)^2-\left(\sqrt{5}\right)^2\right]\)
\(=\left(x+2\right)^2\left(x+3-\sqrt{5}\right)\left(x+3+\sqrt{5}\right)\)
\(\Leftrightarrow-2x+1-x-2=8\cdot\left(-4x^2+6x-2x\right)+4\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow-3x-1+32x^2-48x+16x-4x^2+8x-4=0\)
\(\Leftrightarrow28x^2-27x-5=0\)
\(\text{Δ}=\left(-27\right)^2-4\cdot28\cdot\left(-5\right)=1289>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{27-\sqrt{1289}}{56}\\x_2=\dfrac{27+\sqrt{1289}}{56}\end{matrix}\right.\)
\(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}\)
\(\Leftrightarrow\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}=0\)
\(\Leftrightarrow\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}\right)=0\)
\(\Leftrightarrow x-23=0\left(vì\frac{1}{24}+\frac{1}{25}-\frac{1}{26}\ne0\right)\)
\(\Leftrightarrow x=23\)
vậy................
\(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=0\)
\(\Leftrightarrow\left(\frac{201-x}{99}+1\right)+\left(\frac{203-x}{97}+1\right)+\left(\frac{205-x}{95}+1\right)=0\)
\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)
\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\)
\(\Leftrightarrow300-x=0\left(vì\frac{1}{99}+\frac{1}{97}+\frac{1}{95}>0\right)\)
\(\Leftrightarrow x=300\)
vậy..........
a,A(\(x\)) = 13\(x^4\) + 3\(x^2\) + 15\(x\) - 8\(x\) - 7 - 7\(x\) + 7\(x^2\) - 10\(x^4\)
A(\(x\)) = (13\(x^4\) - 10\(x^4\)) + (3\(x^2\) + 7\(x^2\)) + (15\(x\) - 8\(x\) - 7\(x\)) - 7
A(\(x\)) = 3\(x^4\) + 10\(x^2\) + 0 - 7
A(\(x\)) = 3\(x^4\) + 10\(x^2\) - 7
B(\(x\)) = -4\(x^4\) - 10\(x^2\) + 10 + 5\(x^4\) - 3\(x\) - 18 + 30 - 5\(x^2\)
B(\(x\)) = (-4\(x^4\) + 5\(x^4\)) - (10\(x^2\) + 5\(x^2\)) - 3\(x\) + (10 + 30 - 18)
B(\(x\)) = \(x^4\) - 15\(x^2\) - 3\(x\) + 22
b,C(\(x\)) = A(\(x\)) + B(\(x\)) = 3\(x^4\) + 10\(x^2\) - 7 + \(x^4\) - 15\(x^2\) - 3\(x\) + 22
C(\(x\)) = 4\(x^4\) - (15\(x^2\) - 10\(x^2\)) - 3\(x\) + 22
C(\(x\)) = 4\(x^4\) - 5\(x^2\) - 3\(x\) + 15
c, D(\(x\)) = B(\(x\)) - A(\(x\)) = \(x^4\) - 15\(x^2\) - 3\(x\) + 22 - 3\(x^4\) - 10\(x^2\) + 7
D(\(x\)) = (\(x^4\) - 3\(x^4\)) - (15\(x^2\) + 10\(x^2\)) + (22 + 7)
D(\(x\)) = - 2\(x^4\) - 25\(x^2\) + 29
d, Thay \(x\) = 1 vào C(\(x\)) ta có: C(1) = 4.14 - 5.12 -3.1 + 15 = 11 (xem lại đề bài em nhá)
cái bài 2 câu 1 câu 2 và câu 3 sửa cái vế phải lại thành 3/2-1-2x/4 và -15/5 và 2.(x-1)/5