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Bài 2:
a) \(B=x^2+x-11\)
\(=x^2+2.\frac{1}{2}.x+\frac{1}{4}-\frac{45}{4}\)
\(=\left(x+\frac{1}{2}\right)^2-\frac{45}{4}\ge-\frac{45}{4}\)
Dâu = xảy ra khi:
\(\left(x+\frac{1}{2}\right)^2=0\)
\(\Rightarrow x+\frac{1}{2}=0\)
\(\Rightarrow x=0-\frac{1}{2}=-\frac{1}{2}\)
b) \(C=x^2-8x+1\)
\(=x^2-2.4.x+16-15\)
\(=\left(x-4\right)^2-15\ge-15\)
Dấu = xảy ra khi:
\(\left(x-4\right)^2=0\)
\(\Rightarrow x-4=0\)
\(\Rightarrow x=0+4=4\)
Ko chắc à!
\(a,\left(x+4\right)^2-x\left(x-5\right)=19\)
\(x^2+8x+16-x^2+5x=19\)
\(8x+5x=19-16\)
\(13x=3\)
\(x=\frac{3}{13}\)
\(b,x^2+3x-10=0\)
\(\Rightarrow x^2+5x-2x-10=0\)
\(\Rightarrow x\left(x+5\right)-2\left(x+5\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}}\)
\(a,\left(x+4\right)^2-x\left(x-5\right)=19\)
\(x^2+8x+16-x^2+5x=19\)
\(8x+5x=19-16\)
\(13x=3\)
\(x=\frac{3}{13}\)
\(b,x^2+3x-10=0\)
\(\Rightarrow x^2+5x-2x-10=0\)
\(\Rightarrow x\left(x+5\right)-2\left(x+5\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}}\)
a) (x+4)2 - x(x-5) = 19
x2+2.x.4+42 - x2 +5x = 19
8x +16 +5x =19
13x +16 =19
13x = 19-16=3
=> x=3:13=\(\frac{3}{13}\)
b) x2 +3x -10 =0
x(x+3) -10 =0
x(x+3) =10
=> x=2
chúc bn học tốt nha ^^ t chi mk nhé <3
a) Ta có: \(a\left(m-n\right)+m-n\)
\(=a\left(m-n\right)+\left(m-n\right)\)
\(=\left(m-n\right)\left(a+1\right)\)
b) Ta có: \(mx+my+5x+5y\)
\(=m\left(x+y\right)+5\left(x+y\right)\)
\(=\left(x+y\right)\left(m+5\right)\)
c) Ta có: \(ma+mb-a-b\)
\(=m\left(a+b\right)-\left(a+b\right)\)
\(=\left(a+b\right)\left(m-1\right)\)
d) Ta có: \(1-xa-x+a\)
\(=\left(a+1\right)-x\left(a+1\right)\)
\(=\left(a+1\right)\left(1-x\right)\)
e) Ta có: \(\left(a-b\right)^2-\left(b-a\right)\left(a+b\right)\)
\(=\left(a-b\right)^2+\left(a-b\right)\left(a+b\right)\)
\(=\left(a-b\right)\left(a-b+a+b\right)\)
\(=2a\left(a-b\right)\)
f) Ta có: \(a\left(a-b\right)\left(a+b\right)-\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(=\left(a+b\right)\left(a^2-ab\right)-\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(=\left(a+b\right)\left(a^2-ab-a^2+ab-b^2\right)\)
\(=b^2\cdot\left(a+b\right)\)
g) Ta có: \(3x\left(x+7\right)^2-11x^2\left(x+7\right)+9\left(x+7\right)\)
\(=\left(x+7\right)\left[3x\left(x+7\right)-11x^2+9\right]\)
\(=\left(x+7\right)\left(3x^2+21x-11x^2+9\right)\)
\(=\left(x+7\right)\left(-8x^2+21x+9\right)\)
\(=\left(x+7\right)\left(-8x^2+24x-3x+9\right)\)
\(=\left(x+7\right)\left[-8x\left(x-3\right)-3\left(x-3\right)\right]\)
\(=\left(x+7\right)\left(x-3\right)\left(-8x-3\right)\)
h) Ta có: \(\left(x+5\right)^2-3\left(x+5\right)\)
\(=\left(x+5\right)\left(x+5-3\right)\)
\(=\left(x+5\right)\left(x+2\right)\)
i) Ta có: \(2x\left(x-3\right)-3\left(x-3\right)^2\)
\(=\left(x-3\right)\left[2x-3\left(x-3\right)\right]\)
\(=\left(x-3\right)\left(2x-3x+9\right)\)
\(=\left(x-3\right)\left(9-x\right)\)
j) Ta có: \(x\left(x-7\right)+\left(7-x\right)^2\)
\(=x\left(x-7\right)+\left(x-7\right)^2\)
\(=\left(x-7\right)\left(x+x-7\right)\)
\(=\left(x-7\right)\left(2x-7\right)\)
k) Ta có: \(3x\left(x-9\right)^2-\left(9-x\right)^3\)
\(=3x\left(x-9\right)^2+\left(x-9\right)^3\)
\(=\left(x-9\right)^2\cdot\left(3x+x-9\right)\)
\(=\left(x-9\right)^2\cdot\left(4x-9\right)\)
C1
a) -7x(3x-2)=-21x^2+14x
b) 87^2+26.87+13^2=87^2+2.13.87+13^2=(87+13)^2=100^2
C2
a) (x-5)(x+5)
b)3x(x+5)-2(x+5)=(3x-2)(x+5)=0
\(\Rightarrow\left[\begin{array}{nghiempt}3x-2=0\\x+5=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{2}{3}\\x=-5\end{array}\right.\)
Vậy S={-5;2/3}
C3:
a)3x^3-2x^2+2=(x+1)(3x^2-5x-5)-3
b) Để A chia hết cho B=> x+1\(\inƯ\left(-3\right)\)
\(\Rightarrow\begin{cases}x+1=3\\x+1=-3\\x+1=1\\x+1=-1\end{cases}\)\(\Rightarrow\begin{cases}x=2\\x=-4\\x=0\\x=-2\end{cases}\)
\(\left(5\cdot\left(x^2-3x+1\right)+x\cdot\left(1-5x\right)\right)-\left(x-2\right)=0\)
\(7-15x=0\)
\(-15x=-7\)
\(x=\frac{7}{15}=0.467\)
\(b,\)câu b dài quá nên mik lười, vậy mik ghi kết quả thôi nhé
\(x=\frac{2}{19}=0.105\)
\(c,\)câu c cũng vậy mik ghi kết quả thôi nhé bn
\(x=-\frac{6}{11}=-0.545\)
Bực olm quá! Không cho người ta giải gì hết,cứ giải cần hết bài thì bị bắt tải lại. Nãy giờ hơn 15 lần rồi! Lần nãy nữa không giải nữa đâu nhé olm!!!!! Bực vl!Admin fix nhanh cho em cái! Mấy lần rồi bực quá nên giờ không biết giải còn đúng hay không :v
\(\left(x-5\right)\left(x+5\right)-\left(x+3\right)^2+3\left(x-2\right)^2=\left(x+1\right)^2-\left(x+4\right)\left(x-4\right)+3x^2\)
\(\Leftrightarrow\left(x^2-5^2\right)-\left(x^2+2.3x+3^2\right)+3\left(x^2-2.2x+2^2\right)=\left(x^2+2x+1\right)-\left(x^2-4^2\right)+3x^2\)
\(\Leftrightarrow x^2-25-x^2-6x-9+3x^2-12x+4=x^2+2x+1-x^2+16+3x^2\)
\(\Leftrightarrow\left(x^2-x^2+3x^2\right)-\left(25+9-4\right)-\left(6x+12x\right)=\left(x^2-x^2+3x^2\right)+2x+\left(1+16\right)\)
\(\Leftrightarrow3x^2-30-18x=3x^2+2x+17\)
\(\Leftrightarrow3x^2-3x^2-18x-2x=30+17\)
\(\Leftrightarrow-20x=47\Leftrightarrow x=\frac{-47}{20}\)
1.a) (3x+1)2-4(x-2)2= (3x+1)2-[2(x-2)]2=[(3x+1)-2(x-2)][(3x+1)+2(x-2)]=(x+3)(5x-1)
b) (a2+b2-5)2-4(ab+2)2= (a2+b2-5)2-[2(ab+2)]2 = (a2+b2-5-2ab-4)(a2+b2-5+2ab+4)=[(a-b)2-9][(a+b)2-1]
2. 3x2+9x-30=3x2-6x+15x-30=3x(x-2)+15(x-2)=3(x+5)(x-2)
b. x3-5x2-14x=x3+2x2-7x2-14x=x2(x+2)-7x(x+2)=(x2-7x)(x+2)
a) \(\left(3x+1\right)^2-4\left(x-2\right)^2\)
\(=\left(3x+1\right)^2-\left[2.\left(x-2\right)\right]^2\)
\(=\left(3x+1\right)^2-\left(2x-4\right)^2\)
\(=\left[3x+1-2x+4\right].\left[3x+1+2x-4\right]\)
\(=\left(x+5\right)\left(5x-3\right)\)
b) \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)
\(=\left(a^2+b^2-5\right)^2-\left[2.\left(ab+2\right)\right]^2\)
\(=\left(a^2+b^2-5\right)^2-\left(2ab+4\right)^2\)
\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)
\(=\left[\left(a-b\right)^2-9\right].\left[\left(a+b\right)^2-1\right]\)
\(=\left[\left(a-b-3\right)\left(a-b+3\right)\right].\left[\left(a+b-1\right)\left(a+b+1\right)\right]\)
a) \(3x^2+9x-30\)
\(=3\left(x^2+3x-10\right)\)
\(=3\left(x^2-2x+5x-10\right)\)
\(=3.\left[x\left(x-2\right)+5.\left(x-2\right)\right]\)
\(=3.\left[\left(x+5\right)\left(x-2\right)\right]\)
b) \(x^3-5x^2-14x\)
\(=x\left(x^2-5x-14\right)\)
\(=x\left(x^2+2x-7x-14\right)\)
\(=x.\left[x\left(x+2\right)-7.\left(x+2\right)\right]\)
\(=x.\left[\left(x-7\right)\left(x+2\right)\right]\)
tìm gía trị lớn nhất mà chị