b: \(\Leftrightarrow x-15-27-x+x-13=-1\)

\(\Leftrightarrow x-55=-1\)

hay x=54

Ta có:\(1001=1000+1=x+1\)

\(x^8-1001x^7+1001x^6+...+1001x^2-1001x+250\\ =x^8-\left(x+1\right)x^7+\left(x+1\right)x^6+...+\left(x+1\right)x^2-\left(x+1\right)x\\ =x^8-x^8-x^7+x^7+x^6+...+x^3+x^2-x^2-x+250\\ =-x+250=-1000+250\\ =-750\)

oa

a) Câu hỏi của Hinamori Amu - Toán lớp 7 | Học trực tuyến

a,+b,:Câu hỏi của Hinamori Amu - Toán lớp 7 | Học trực tuyến

a) M=\(x^3+x^2y-xy^2-y^3+x^2-y^2+2x+2y+3\)

=\(x^2\left(x+y+1\right)-y^2\left(x+y+1\right)+2\left(x+y+1\right)+1\)

=\(x^2.0-y^2.0+2.0+1=1\)

Vậy với x+y+1=0 thì M=1

b) hình như thiếu đề

lấy:1/1001x1001=1

nhầm nhé 1001.1/1001=1

Vì có 1001 phân số 1/1001 cộng với nhau nên

1001+1/1001=bằng 1001,000999

\(A.x=x+x^2+x^3+...+x^{101}\)

\(A.x-A=x^{101}-1\Rightarrow A\left(x-1\right)=x^{101}-1\)

\(\Rightarrow A=\dfrac{x^{101}-1}{x-1}\)

5=1001−𝑥+1001

5=2002−𝑥

5=−𝑥+2002

5−2002=−𝑥+2002−2002

−1997=−𝑥

𝑥=1997

\(-16+8+13=1001-x+1001\)

⇔\(5=2002-x\)

⇔\(x=1997\)

\(\text{- 16 + 8 + 13 = 1001 - x + 1001}\\ 5=1001-x+1001\\ 5=2002-x\\ 2002-x=5\\ x=2002-5\\ x=1997\)

\(\Leftrightarrow\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\)

\(\Leftrightarrow\frac{x^4b+y^4a}{ab}=\frac{x^4+y^4+2x^2y^2}{a+b}\Leftrightarrow\left(a+b\right)\left(x^4b+y^4a\right)=ab\left(x^4+y^4+2x^2y^2\right)\)

\(\Leftrightarrow x^4ab+y^4a^2+x^4b^2+y^4ab=x^4ab+y^4ab+2x^2y^2ab\)

\(\Leftrightarrow y^4a^2+x^4b^2=2x^2y^2ab\Leftrightarrow\left(x^2b-y^2a\right)^2=0\Leftrightarrow\frac{x^2}{a}=\frac{y^2}{b}\)

\(\Rightarrow\left(\frac{x^2}{a}\right)^{1001}=\left(\frac{y^2}{b}\right)^{1001}\Leftrightarrow\frac{x^{2002}}{a^{1001}}=\frac{y^{2002}}{b^{2011}}\)

Mà: \(\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\Leftrightarrow\left(\frac{x^2}{a}\right)^{1001}=\frac{1}{\left(a+b\right)^{1001}}\)

\(\Rightarrow\frac{x^{2002}}{a^{1001}}+\frac{y^{2002}}{b^{1001}}=\frac{2}{\left(a+b\right)^{1001}}\left(đpcm\right)\)

\(x^2+y^2=1\Rightarrow y^2=1-x^2\)

\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\Leftrightarrow\frac{b.x^4+a.y^4}{ab}=\frac{1}{a+b}\)

\(\Leftrightarrow bx^4+ay^4=\frac{ab}{a+b}\Leftrightarrow bx^4+a\left(1-x^2\right)^2-\frac{ab}{a+b}=0\)

\(\Leftrightarrow bx^4+a\left(x^4-2x^2+1\right)-\frac{ab}{a+b}=0\)

\(\Leftrightarrow\left(a+b\right)x^4-2ax^2+a-\frac{ab}{a+b}=0\)

\(\Leftrightarrow\left(a+b\right)x^4-2ax^2+\frac{a^2}{a+b}=0\Leftrightarrow\left(a+b\right)\left[x^4-2.x.\frac{a}{a+b}+\left(\frac{a}{a+b}\right)^2\right]=0\)

\(\Leftrightarrow\left(a+b\right)\left(x^2-\frac{a}{a+b}\right)=0\Rightarrow x^2=\frac{a}{a+b}\) (do \(a+b\ne0\))

\(\Rightarrow y^2=1-x^2=\frac{b}{a+b}\)

\(\Rightarrow\) \(\frac{x^2}{a}=\frac{a}{a\left(a+b\right)}=\frac{1}{a+b}\) ; \(\frac{y^2}{b}=\frac{b}{b\left(a+b\right)}=\frac{1}{a+b}\)

Thay vào bài toán:

\(\frac{x^{2002}}{a^{1001}}+\frac{y^{2002}}{b^{1001}}=\left(\frac{x^2}{a}\right)^{1001}+\left(\frac{y^2}{b}\right)^{1001}=\left(\frac{1}{a+b}\right)^{1001}+\left(\frac{1}{a+b}\right)^{1001}=\frac{2}{\left(a+b\right)^{1001}}\)

$x^2+y^2=1\Rightarrow y^2=1-x^2$

$\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\iff \frac{b.x^4+a.y^4}{ab}=\frac{1}{a+b}$

$\iff b{x}^4+a{y}^4=\frac{ab}{a+b}\iff b{x}^4+a{\left(1-x^2\right)}^2-\frac{ab}{a+b}=0$

$\iff b{x}^4+a\left(x^4-2x^2+1\right)-\frac{ab}{a+b}=0$

$\iff \left(a+b\right)x^4-2a{x}^2+a-\frac{ab}{a+b}=0$

$\iff \left(a+b\right)x^4-2a{x}^2+\frac{a^2}{a+b}=0\iff \left(a+b\right)\left[x^4-2.x.\frac{a}{a+b}+{\left(\frac{a}{a+b}\right)}^2\right]=0$

$\iff \left(a+b\right)\left(x^2-\frac{a}{a+b}\right)=0\Rightarrow x^2=\frac{a}{a+b}$ (do $a+b\ne 0$)

$\Rightarrow y^2=1-x^2=\frac{b}{a+b}$

$\Rightarrow$ $\frac{x^2}{a}=\frac{a}{a\left(a+b\right)}=\frac{1}{a+b}$ ; $\frac{y^2}{b}=\frac{b}{b\left(a+b\right)}=\frac{1}{a+b}$

Thay

$\frac{x^{2002}}{a^{1001}}+\frac{y^{2002}}{b^{1001}}={\left(\frac{x^2}{a}\right)}^{1001}+{\left(\frac{y^2}{b}\right)}^{1001}={\left(\frac{1}{a+b}\right)}^{1001}+{\left(\frac{1}{a+b}\right)}^{1001}=\frac{2}{{\left(a+b\right)}^{1001}}$