51/30

24/65

chưa xong 

Em nên gõ công thức trực quan để đề bài rõ ràng nhé

câu 2 phần 2:

\(\left\{{}\begin{matrix}4x+3y=11\\4x-y=7\end{matrix}\right.\)\(< =>\left\{{}\begin{matrix}4y=4\\4x-y=7\end{matrix}\right.< =>\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\).Vậy hệ pt có nghiệm

(x,y)=(2;1)

caau3 phần 2:

\(x^2-2x+m-1=0\)(1)

\(\Delta'=\left(-1\right)^2-\left(m-1\right)=1-m+1=2-m\)

để pt (1) có 2 nghiệm x1,x2<=>\(\Delta'\ge0< =>2-m\ge0< =>m\le2\)

theo vi ét=>\(\left\{{}\begin{matrix}x1+x2=2\left(1\right)\\x1.x2=m-1\left(3\right)\end{matrix}\right.\)

có: \(x1^4\)\(-x1^3=x2^4-x2^3\)

\(< =>x1^4-x2^4-x1^3+x2^3=0\)

\(< =>\left(x1^2-x2^2\right)\left(x1^2+x2^2\right)-\left(x1^3-x2^3\right)\)\(=0\)

\(< =>\left(x1-x2\right)\left(x1+x2\right)\left[\left(x1+x2\right)^2-2x1x2\right]\)\(-\left(x1-x2\right)\left(x1^2+x1x2+x^2\right)=0\)

\(< =>\)\(\left(x1-x2\right)\left[2.2^2-2\left(m-1\right)-\left(x1^2+x1x2+x2^2\right)\right]=0\)

\(< =>.\left(x1-x2\right)\left[8-2m+2-\left(x1+x2\right)^2+x1x2\right]=0\)

<=>\(\left(x1-x2\right)\left[10-2m-4+m-1\right]=0\)

\(< =>\left(x1-x2\right)\left(5-m\right)=0\)

\(=>\left[{}\begin{matrix}x1-x2=0\\5-m=0\end{matrix}\right.< =>\left[{}\begin{matrix}x1=x2\left(2\right)\\m=5\left(loai\right)\end{matrix}\right.\)

thế(2) vào(1)=>\(x1=x2=1\left(4\right)\)

thế (4) vào (3)=>\(m-1=1=>m=2\left(TM\right)\)

vậy m=2 thì....

câu 2: b

câu 3: c

câu 4: 3 phần 4

 b

c

3/4

b: Ta có: \(a=\sqrt{\dfrac{5}{2}-\sqrt{6}}\)

\(=\dfrac{\sqrt{5-2\sqrt{6}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{2}}\)

Ta có: \(P=1+\sqrt{a}\)

\(=1+\sqrt{\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{2}}}\)

\(=1+\sqrt{\dfrac{2\sqrt{6}-4}{4}}\)

\(=1+\dfrac{\sqrt{2\sqrt{6}-4}}{2}\)

\(=\dfrac{2+\sqrt{2\sqrt{6}-4}}{2}\)

C1 19991999 x 1998 ‐ 19981998 x 1999
= 1999 x 10001 x 1998 ‐ 1998 x 10001 x 1999
= 0

C2 ﴾ 1999 x 1998 + 1998 x 1997 ﴿ x ﴾ 1 + 1/3 ‐ 11/3 ﴿
= ﴾ 1999 x 1998 + 1998 x 1997 ﴿ x ﴾ 11/3 ﴾Hỗn số﴿ ‐ 11/3﴿
= ﴾ 1999 x 1998 + 1998 x 1997﴿ x 0
= 0

giup mk voi 1999. 1999. 1998- 1998

Sao toàn viết chữ thế ? Đọc sao nổi ? Viết lại ra số đi

\(a.\frac{2}{3}+\frac{1}{5}\cdot\frac{10}{7}\)

\(=\frac{2}{3}+\frac{1\cdot2}{1\cdot7}\)

\(=\frac{2}{3}+\frac{2}{7}=\frac{2\cdot7}{21}+\frac{2\cdot3}{21}=\frac{14}{21}+\frac{6}{21}=\frac{20}{21}\)

\(b.\frac{2}{7}\cdot\frac{4}{7}+\frac{2}{7}\cdot\frac{3}{7}\)

\(=\frac{2}{7}\cdot\left(\frac{4}{7}+\frac{3}{7}\right)\)

\(=\frac{2}{7}\cdot\frac{7}{7}=\frac{2}{7}\cdot1=\frac{2}{7}\)

\(c.\left[-\frac{1}{4}+\frac{3}{10}\right]:\left(-\frac{3}{5}\right)-\frac{7}{6}\)

\(=\left[-\frac{5}{20}+\frac{6}{20}\right]:\left(-\frac{3}{5}\right)-\frac{7}{6}\)

\(=\frac{1}{20}:\left(-\frac{3}{5}\right)-\frac{7}{6}\)

\(=\frac{1}{20}\cdot\left(-\frac{5}{3}\right)-\frac{7}{6}\)

\(=\frac{1\cdot\left(-1\right)}{4\cdot3}-\frac{7}{6}\)

\(=\left(-\frac{1}{12}\right)-\frac{7}{6}=\left(-\frac{1}{12}\right)-\frac{14}{12}=-\frac{15}{12}\)

bài 2 :a) \(2x-2\frac{2}{7}=2\frac{5}{7}\)

\(2x=2\frac{5}{7}-2\frac{2}{7}\)

\(2x=\left(2-2\right)+\left(\frac{5}{7}-\frac{2}{7}\right)\)

\(2x=0+\frac{3}{7}\)

\(2x=\frac{3}{7}\)

\(x=\frac{3}{7}:2=\frac{3}{7}\cdot\frac{1}{2}=\frac{3}{14}\)

\(b.\frac{17}{13x}=\frac{4}{39}\)

\(\Rightarrow13x\cdot4=17\cdot39\)

\(\Rightarrow13x\cdot4=663\)

\(\Rightarrow13x=663:4\)

\(\Rightarrow13x=165,75\)

\(\Rightarrow x=165,75:13\)

\(\Rightarrow x=12,75\)

\(****nha!!!!!!!!!!!!\)

dung may inh la tinh ra 

Câu 1: C

Câu 2: A

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