a: (x+2)(x-3)>0

nên x+2;x-3 cùng dấu

=>x>3 hoặc x<-2

b: (x-1)(x+4)<=0

nên x-1 và x+4 khác dấu

=>-4<=x<=1

\(f\left(x\right)-g\left(x\right)=\left(x^5-3x^2+x^3-x^2-2x+5\right)-\left(x^2-3x+1+x^2-x^4+x^5\right)\)

\(f\left(x\right)-g\left(x\right)=x^5-3x^2+x^3-x^2-2x+5-x^2+3x-1-x^2+x^4-x^5\)

\(f\left(x\right)-g\left(x\right)=\left(x^5-x^5\right)+\left(-3x^2-x^2-x^2-x^2\right)+x^3+\left(-2x+3x\right)+\left(5-1\right)+x^4\)

\(f\left(x\right)-g\left(x\right)=-6x^2+x^3+x+4+x^4\)

\(f\left(x\right)-g\left(x\right)=x^4+x^3-6x^2+x+4\)

Ta có: x=2 

nên x-1=1

Ta có: \(B=\left(x+1\right)\left(x^7-x^6+x^5-x^4+x^3-x^2+x-1\right)\)

\(=\left(x+1\right)\left[x^6\left(x-1\right)+x^4\left(x-1\right)+x^2\left(x-1\right)+\left(x-1\right)\right]\)

\(=\left(x+1\right)\left(x^6+x^4+x^2+1\right)\)

\(=\left(x+1\right)\left(x+1\right)\left(x^4+1\right)\)

\(=\left(2^4+1\right)\left(2+1\right)^2=17\cdot9=153\)

bằng 101606400 à? có đúng không? nếu không thì k cho mình là sai cũng được nhé.

= 101606400

kq như z k bt đúng hay sai nha

1: \(\frac{1}{4+2\sqrt2}+\frac{1}{4-2\sqrt2}\)

\(=\frac{4-2\sqrt2+4+2\sqrt2}{\left(4-2\sqrt2\right)\left(4+2\sqrt2\right)}=\frac{8}{16-8}=1\)

2: \(\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}-\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}\)

\(=\frac{\left(\sqrt5+\sqrt3\right)^2-\left(\sqrt5-\sqrt3\right)^2}{\left(\sqrt5+\sqrt3\right)\left(\sqrt5-\sqrt3\right)}=\frac{8+2\sqrt{15}-\left(8-2\sqrt{15}\right)}{5-3}\)

\(=\frac{4\sqrt{15}}{2}=2\sqrt{15}\)

3: \(\left(\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}+1\right):\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}=\frac{\sqrt5-\sqrt3+\sqrt5+\sqrt3}{\sqrt5+\sqrt3}\cdot\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}\)

\(=\frac{2\sqrt5}{\sqrt5-\sqrt3}=\frac{2\sqrt5\left(\sqrt5+\sqrt3\right)}{5-3}=\sqrt5\left(\sqrt5+\sqrt3\right)=5+\sqrt{15}\)

4: \(\sqrt{\left(1-\sqrt3\right)^2}+\frac{1}{\sqrt{\left(1+\sqrt3\right)^2}}\)

\(=\sqrt3-1+\frac{1}{\sqrt3+1}=\frac{\left(\sqrt3-1\right)\left(\sqrt3+1\right)+1}{\sqrt3+1}=\frac{3-1+1}{\sqrt3+1}\)

\(=\frac{3\left(\sqrt3-1\right)}{2}\)

5: \(\frac{3}{\sqrt7-2}-\frac{3}{\sqrt7+2}=\frac{3\left(\sqrt7+2\right)-3\left(\sqrt7-2\right)}{\left(\sqrt7-2\right)\left(\sqrt7+2\right)}\)

\(=\frac{3\sqrt7+6-3\sqrt7+6}{7-4}=\frac{12}{3}=4\)

6: \(\frac{5\sqrt2-2\sqrt5}{\sqrt5-\sqrt2}-\frac{9}{\sqrt{10}+1}\)

\(=\frac{\sqrt{10}\left(\sqrt5-\sqrt2\right)}{\sqrt5-\sqrt2}-\frac{9\left(\sqrt{10}-1\right)}{10-1}=\sqrt{10}-\left(\sqrt{10}-1\right)=1\)

7: \(\frac{2\sqrt3}{\sqrt3+\sqrt2}+\sqrt{24}\)

\(=\frac{2\sqrt3\left(\sqrt3-\sqrt2\right)}{\left(\sqrt3-\sqrt2\right)\left(\sqrt3+\sqrt2\right)}+2\sqrt6\)

\(=2\sqrt3\left(\sqrt3-\sqrt2\right)+2\sqrt6=6-2\sqrt6+2\sqrt6=6\)

8: \(\sqrt3\left(2\sqrt{27}-\sqrt{75}+\frac32\cdot\sqrt{12}\right)\)

\(=\sqrt3\left(2\cdot3\sqrt3-5\sqrt3+\frac32\cdot2\sqrt3\right)=\sqrt3\left(6\sqrt3-5\sqrt3+3\sqrt3\right)=\sqrt3\cdot4\sqrt3=12\)

9: \(\frac{\sqrt{8-2\sqrt{12}}}{\sqrt3-1}=\frac{\sqrt{\left(\sqrt6-\sqrt2\right)^2}}{\sqrt3-1}=\frac{\sqrt2\left(\sqrt3-1\right)}{\sqrt3-1}=\sqrt2\)

b) \(B=\sqrt{12+2\sqrt{35}}=\sqrt{12+2.\sqrt{7}.\sqrt{5}}=\sqrt{\left(\sqrt{7}\right)^2+2.\sqrt{7}.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{7}+\sqrt{5}\right)^2}=\left|\sqrt{7}+\sqrt{5}\right|\)

Vì \(\sqrt{7}>\sqrt{5}\) nên \(\left|\sqrt{7}+\sqrt{5}\right|=\sqrt{7}+\sqrt{5}\)

d) \(D=\sqrt{6-\sqrt{35}}+\sqrt{10}=\sqrt{2}.\sqrt{6-\sqrt{35}}+\sqrt{5}=\sqrt{12-\sqrt{35}}+\sqrt{5}=\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}+\sqrt{5}=\left|\sqrt{7}-\sqrt{5}\right|+\sqrt{5}=\sqrt{7}-\sqrt{5}+\sqrt{5}=\sqrt{7}\)

@Khuynh ly Nam cung

Bạn bấm @ xong bạn gõ tên ra, bạn sẽ thấy ở góc của hộp thoại có tên bạn ấy và avt của bạn ấy rồi bạn bấm vào nha

Ủa để làm gì ?

x₁;x₂;x₃;x₄;x₅=-1 hoặc 1

=>x₁.x₂;x₂.x₃;x₃.x₄;x₄.x₅;x₅.x₁ bằng 1 hoặc -1

giả sử x₁.x₂+x₂.x₃+x₃.x₄+x₄.x₅+x₅.x₁=0

=>số các số hạng 1 và -1 bằng nhau

=>số các số hạng chia hết cho 2

=>5 chia hết cho 2(có 5 số hạng) Vô lí

=>x₁.x₂+x₂.x₃+x₃.x₄+x₄.x₅+x₅.x₁\(\ne0\)

=>đpcm

chtt

ai làm ơn tích mình ,mình tích lại cho