68 : 2< ....................<36
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1) \(2x\left(x-5\right)+\left(x-2\right)\left(x+3\right)=2x^2-10x+x^2+3x-2x-6=3x^2-9x-6\)
2) \(\left(2x-5\right)\left(1-x\right)-\left(x-3\right)\left(-2x\right)=2x-2x^2-5+5x+2x^2-6x=x-5\)
3) \(\left(4x-3\right)\left(4x-3\right)-\left(3x+2\right)\left(3x-2\right)=\left(4x-3\right)^2-9x^2+4=16x^2-24x+9-9x^2+4\)
\(=7x^2-24x+13\)
4) \(\left(2x-1\right)\left(2x+1\right)\left(2x+1\right)-4\left(x^2+1\right)=\left(2x-1\right)[\left(2x+1\right)^2]-4x^2-4\)
\(=\left(2x-1\right)\left(4x^2+4x+4\right)-4x^2-4=8x^3+8x^2+8x-4x^2-4x-4-4x^2-4=8x^3+4x-8\)
5) \(3x\left(2x-8\right)-\left(2-6x\right)\left(5+x\right)=6x^2-24x-10-2x+30x+6x^2=12x^2+4x-10\)
6) \(x\left(3x-18\right)-3\left(x-4\right)\left(x-2\right)+8=3x^2-18x-3x^2+6x+12x-24+8=-16\)
7) \(\left(x+2\right)\left(x^2-2x+4\right)-x^2\left(x-2\right)-2x^2=x^3+8-x^3+2x^2-2x^2=8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
TA CÓ: Vi 2\(⋮\)2 nên A chia het cho 2
Vi 2+\(2^2\)=6 \(⋮\)3 nên A chia het cho3
Vi tớ ko biet nua sao chia het cho 5 dc nhi
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Delta'=\left(k-1\right)^2+4k=k^2+2k+1=\left(k+1\right)^2\ge0;\forall k\)
\(\Rightarrow\) Phương trình luôn có nghiệm với mọi k
b. Để phương trình có 2 nghiệm pb \(\Rightarrow k\ne-1\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(k-1\right)\\x_1x_2=-4k\end{matrix}\right.\)
Kết hợp với điều kiện đề bài ta được:
\(\left\{{}\begin{matrix}x_1+x_2=2\left(k-1\right)\\3x_1-x_2=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x_1+x_2=2k-2\\4x_1=2k\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{k}{2}\\x_2=\dfrac{3k-4}{2}\end{matrix}\right.\)
Thế vào \(x_1x_2=-4k\)
\(\Rightarrow\dfrac{k}{2}.\left(\dfrac{3k-4}{2}\right)=-4k\)
\(\Leftrightarrow3k^2-4k=-16k\)
\(\Leftrightarrow3k^2+12k=0\Rightarrow\left[{}\begin{matrix}k=0\\k=-4\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
theo vi ét ta có S=m+1 và P=m-4
bạn nhân tung pt (x12 - mx1 + m)(x22 - mx2 + m) = 2 sẽ được
\(x1^2x2^2-mx1x2\left(x1+x2\right)+m^2x1x2-m^2\left(x1+x2\right)+m\left(x1^2+x2^2\right)+m^2\)
lưu ý \(x1^2+x2^2=\left(x1+x2\right)^2-2x1x2\)
bạn thay S=x1+x2 và P=x1x2 vào rồi giải pt ẩn m là ra
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^2-6x-17=\left(x^2-6x+9\right)-26=\left(x-3\right)^3-26\ge-26\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-3\right)^2=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
b)\(x^2-10x=\left(x^2-10x+25\right)-25=\left(x-5\right)^2-25\ge-25\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Leftrightarrow x-5=0\)
\(\Leftrightarrow x=5\)
c)\(3x^2-12x+5=3\left(x^2-4x+4\right)-7=3\left(x-2\right)^2-7\ge-7\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
d)\(2x^2-x+1=2\left(x^2-\frac{x}{2}+\frac{1}{16}\right)+\frac{7}{8}=2\left(x-\frac{1}{4}\right)^2+\frac{7}{8}\ge\frac{7}{8}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-\frac{1}{4}\right)^2=0\)
\(\Leftrightarrow x-\frac{1}{4}=0\)
\(\Leftrightarrow x=\frac{1}{4}\)
e)\(x^2+y^2-8x+4y+27=\left(x^2-8x+16\right)+\left(y^2+4y+4\right)+7=\left(x-4\right)^2+\left(y+2\right)^2+7\ge7\forall x\)Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-4\right)^2=0\\\left(y+2\right)^2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x-4=0\\y+2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=4\\y=-2\end{cases}}\)
f)\(x\left(x-6\right)=x^2-6x=\left(x^2-6x+9\right)-9=\left(x-3\right)^2-9\ge-9\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-3\right)^2=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
h)\(\left(x-2\right)\left(x-5\right)\left(x^2-7x+10\right)=\left(x-2\right)\left(x-5\right)\left(x-2\right)\left(x-5\right)=\left[\left(x-2\right)\left(x-5\right)\right]^2\ge0\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-2\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=5\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1: Thực hiện phép tính và phân tích kết quả ra thừa số nguyên tố
a) \(160-\left(2^3\times5^2-6.25\right)\)
\(=160-\left(8.25-150\right)\)
\(=160-\left(200-150\right)\)
\(=160-50\)
\(=110\)
\(110=2.5.11\)
b) \(4\times5^2-32\div2^4\)
\(=4\times25-32\div16\)
\(=100-2\)
\(=98\)
\(98=2.7^2\)
c) \(5871\div\left[928-\left(247-82.5\right)\right]\)
\(=5871\div\left[928-\left(247-410\right)\right]\)
\(=5871\div\left[928-\left(-163\right)\right]\)
\(=5871\div\left[928+163\right]\)
\(=5871\div1091\)
\(=\frac{5871}{1091}\)
d) \(777\div7+1331\div11^3\)
\(=111+1331\div1331\)
\(=111+1\)
\(=112\)
\(112=2^4.7\)
e) \(6^2\div4.3+2.5^2\)
\(=12\div4.3+2.25\)
\(=9+50\)
\(=59\)
g) \(5.4^2-18\div3^2\)
\(=5.16-18\div9\)
\(=80-2\)
\(=78\)
\(78=2.3.13\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 1/2=5/10; 3/5=6/10; 2/10
b)2/3=8/12; 5/12; 3/4=9/12
~HT~
a) Mẫu số chung: 10
Ta có: \(\frac{1}{2}=\frac{1\times5}{2\times5}=\frac{5}{10}\) ; \(\frac{3}{5}=\frac{3\times2}{5\times2}=\frac{6}{10}\) ; \(\frac{2}{10}\) (giữ nguyên)
\(\Rightarrow\)Ta được những phân số sau: \(\frac{5}{10};\frac{6}{10};\frac{2}{10}\)
b) Mẫu số chung: 12
Ta có: \(\frac{2}{3}=\frac{2\times4}{3\times4}=\frac{8}{12}\) ; \(\frac{5}{12}\) ( giữ nguyên ) ; \(\frac{3}{4}=\frac{3\times3}{4\times3}=\frac{9}{12}\)
\(\Rightarrow\)Ta được những phân số sau: \(\frac{8}{12};\frac{5}{12};\frac{9}{12}\)
@Duongg
68 : 2 = 34 < 35 < 36
68 : 2 = 34 < 35 < 36