Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
a) Ta có: \(VT=\frac{-u^2+3u-2}{\left(u+2\right)\left(u-1\right)}\)
\(=\frac{-\left(u^2-3u+2\right)}{\left(u+2\right)\left(u-1\right)}\)
\(=\frac{-\left(n^2-u-2u+2\right)}{\left(u+2\right)\left(u-1\right)}\)
\(=\frac{-\left[u\left(u-1\right)-2\left(u-1\right)\right]}{\left(u+2\right)\left(u-1\right)}\)
\(=\frac{-\left(u-1\right)\left(u-2\right)}{\left(u+2\right)\left(u-1\right)}\)
\(=\frac{2-u}{u+2}\)(1)
Ta có: \(VP=\frac{u^2-4u+4}{4-u^2}\)
\(=\frac{\left(u-2\right)^2}{-\left(u-2\right)\left(u+2\right)}\)
\(=\frac{-\left(u-2\right)}{u+2}\)
\(=\frac{2-u}{u+2}\)(2)
Từ (1) và (2) suy ra \(\frac{-u^2+3u-2}{\left(u+2\right)\left(u-1\right)}=\frac{u^2-4u+4}{4-u^2}\)
b) Ta có: \(VT=\frac{v^3+27}{v^2-3v+9}\)
\(=\frac{\left(v+3\right)\left(v^3-3u+9\right)}{v^2-3u+9}\)
\(=v+3=VP\)(đpcm)
Bài 2:
a) Ta có: \(\frac{3x^2-2x-5}{M}=\frac{3x-5}{2x-3}\)
\(\Leftrightarrow\frac{3x^2-5x+3x-5}{M}=\frac{3x-5}{2x-3}\)
\(\Leftrightarrow\frac{x\left(3x-5\right)+\left(3x-5\right)}{M}=\frac{3x-5}{2x-3}\)
\(\Leftrightarrow\frac{\left(3x-5\right)\left(x+1\right)}{M}=\frac{3x-5}{2x-3}\)
\(\Leftrightarrow M=\frac{\left(3x-5\right)\left(x+1\right)\left(2x-3\right)}{3x-5}\)
\(\Leftrightarrow M=\left(x+1\right)\left(2x-3\right)\)
\(\Leftrightarrow M=2x^2-3x+2x-3\)
hay \(M=2x^2-x-3\)
Vậy: \(M=2x^2-x-3\)
b) Ta có: \(\frac{2x^2+3x-2}{x^2-4}=\frac{M}{x^2-4x+4}\)
\(\Leftrightarrow\frac{2x^2+4x-x-2}{\left(x-2\right)\left(x+2\right)}=\frac{M}{\left(x-2\right)^2}\)
\(\Leftrightarrow\frac{2x\left(x+2\right)-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{M}{\left(x-2\right)^2}\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(2x-1\right)}{\left(x+2\right)\left(x-2\right)}=\frac{M}{\left(x-2\right)^2}\)
\(\Leftrightarrow\frac{M}{\left(x-2\right)^2}=\frac{2x-1}{x-2}\)
\(\Leftrightarrow M=\frac{\left(2x-1\right)\left(x-2\right)^2}{\left(x-2\right)}\)
\(\Leftrightarrow M=\left(2x-1\right)\left(x-2\right)\)
\(\Leftrightarrow M=2x^2-4x-x+2\)
hay \(M=2x^2-5x+2\)
Vậy: \(M=2x^2-5x+2\)
Bài 3:
a) Ta có: \(\frac{x+1}{N}=\frac{x^2-2x+4}{x^3+8}\)
\(\Leftrightarrow\frac{x+1}{N}=\frac{x^2-2x+4}{\left(x+2\right)\left(x^2-2x+4\right)}\)
\(\Leftrightarrow\frac{x+1}{N}=\frac{1}{x+2}\)
\(\Leftrightarrow N=\left(x+1\right)\left(x+2\right)\)
hay \(N=x^2+3x+2\)
Vậy: \(N=x^2+3x+2\)
n) Ta có: \(\frac{\left(x-3\right)\cdot N}{3+x}=\frac{2x^3-8x^2-6x+36}{2+x}\)
\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=\frac{2x^3+4x^2-12x^2-24x+18x+36}{x+2}\)
\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{\left(x+3\right)}=\frac{2x^2\left(x+2\right)-12x\left(x+2\right)+18\left(x+2\right)}{x+2}\)
\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=\frac{\left(x+2\right)\left(2x^2-12x+18\right)}{x+2}\)
\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=2x^2-12x+18\)
\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=2x^2-6x-6x+18=2x\left(x-3\right)-6\left(x-3\right)=2\cdot\left(x-3\right)^2\)
\(\Leftrightarrow N\cdot\left(x-3\right)=\frac{2\left(x-3\right)^2}{x+3}\)
\(\Leftrightarrow N=\frac{2\left(x-3\right)^2}{x+3}:\left(x-3\right)=\frac{2\left(x-3\right)^2}{\left(x+3\right)\left(x-3\right)}\)
\(\Leftrightarrow N=\frac{2\left(x-3\right)}{x+3}\)
hay \(N=\frac{2x-6}{x+3}\)
Vậy: \(N=\frac{2x-6}{x+3}\)
Câu 1:
$x-2=0\Leftrightarrow 2(x-2)=0\Leftrightarrow 2x-4=0$
Đáp án B.
Câu 2:
Để PT đã cho có nghiệm $x=5$ thì $m(5-3)=6$
$\Leftrightarrow m=3$
Đáp án C
Câu 3:
Dựa vào khái niệm pt bậc nhất 1 ẩn. Đáp án B
Câu 4:
$2x-4=0\Leftrightarrow \frac{2x-4}{4}=0\Leftrightarrow \frac{x}{2}-1=0$
Đáp án D
a) \(\frac{5}{2x+6}\) và \(\frac{7}{12x^3y^4}\)
Ta có: \(2x+6=2\left(x+3\right)\)
\(12x^3y^4=12x^3y^4\)
\(MSC=12x^3y^4\left(x+3\right)\)
Ta có: \(\frac{5}{2x+6}=\frac{5}{2\left(x+3\right)}=\frac{5\cdot6\cdot x^3y^4}{2\cdot6\cdot x^3y^4\cdot\left(x+3\right)}=\frac{30x^3y^4}{12x^3y^4\left(x+3\right)}\)
\(\frac{7}{12x^3y^4}=\frac{7\cdot\left(x+3\right)}{12x^3y^4\cdot\left(x+3\right)}=\frac{7x+21}{12x^3y\left(x+3\right)}\)
b)\(\frac{4}{15x^3y^5}\) và \(\frac{11}{12x^4y^2}\)
MSC=\(60x^4y^5\)
Ta có: \(\frac{4}{15x^3y^5}=\frac{4\cdot4\cdot x}{15x^3y^5\cdot4\cdot x}=\frac{16x}{60x^4y^5}\)
\(\frac{11}{12x^4y^2}=\frac{11\cdot5\cdot y^3}{12x^4y^2\cdot5\cdot y^3}=\frac{55y^3}{60x^4y^5}\)
c) \(\frac{5}{2x+6}\) và \(\frac{3}{x^2-9}\)
Ta có: \(2x+6=2\left(x+3\right)\)
\(x^2-9=\left(x-3\right)\left(x+3\right)\)
MSC=2(x+3)(x-3)
Ta có: \(\frac{5}{2x+6}=\frac{5}{2\left(x+3\right)}=\frac{5\cdot\left(x-3\right)}{2\left(x+3\right)\left(x-3\right)}=\frac{5x-15}{2\left(x+3\right)\left(x-3\right)}\)
\(\frac{3}{x^2-9}=\frac{3}{\left(x-3\right)\left(x+3\right)}=\frac{3\cdot2}{2\cdot\left(x-3\right)\cdot\left(x+3\right)}=\frac{6}{2\left(x-3\right)\left(x+3\right)}\)
d) \(\frac{2x}{x^2-8x+16}\) và \(\frac{x}{3x^2-12x}\)
Ta có: \(x^2-8x+16=\left(x-4\right)^2\)
\(3x^2-12x=3x\left(x-4\right)\)
MSC=\(3x\left(x-4\right)^2\)
Ta có: \(\frac{2x}{x^2-8x+16}=\frac{2x}{\left(x-4\right)^2}=\frac{2x\cdot3x}{3x\cdot\left(x-4\right)^2}=\frac{6x^2}{3x\left(x-4\right)^2}\)
\(\frac{x}{3x^2-12x}=\frac{x}{3x\left(x-4\right)}=\frac{x\left(x-4\right)}{3x\left(x-4\right)^2}=\frac{x^2-4x}{3x\left(x-4\right)^2}\)
e) \(\frac{4x^2-3x+5}{x^3-1}\); \(\frac{1-2x}{x^2+x+1}\) và -2
Ta có: \(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)
MSC=\(\left(x-1\right)\left(x^2+x+1\right)\)
Ta có: \(\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-2x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{1-2x}{x^2+x+1}=\frac{\left(1-2x\right)\left(x-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}=\frac{3x-2x^2-1}{\left(x^2+x+1\right)\left(x-1\right)}\)
\(-2=\frac{-2\left(x^2+x+1\right)\left(x-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}=\frac{-2\left(x^3-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}=\frac{-2x^3+2}{\left(x^2+x+1\right)\left(x-1\right)}\)
f) \(\frac{10}{x+2}\) và \(\frac{5}{2x-4}\) và \(\frac{1}{6-3x}\)
Ta có: \(x+2=x+2\)
\(2x-4=2\left(x-2\right)\)
\(6-3x=3\left(2-x\right)=-3\left(x-2\right)\)
MSC=-6(x-2)(x+2)
Ta có: \(\frac{10}{x+2}=\frac{10\cdot\left(-6\right)\cdot\left(x-2\right)}{\left(x+2\right)\cdot\left(-6\right)\cdot\left(x-2\right)}=\frac{-60\left(x-2\right)}{-6\left(x-2\right)\left(x+2\right)}=\frac{120-60x}{-6\left(x-2\right)\left(x+2\right)}\)
\(\frac{5}{2x-4}=\frac{5}{2\left(x-2\right)}=\frac{5\cdot\left(-3\right)\cdot\left(x+2\right)}{2\cdot\left(x-2\right)\cdot\left(-3\right)\cdot\left(x+2\right)}=\frac{-15\left(x+2\right)}{-6\left(x-2\right)\left(x+2\right)}=\frac{-15x-30}{-6\left(x-2\right)\left(x+2\right)}\)
\(\frac{1}{6-3x}=\frac{1}{3\left(2-x\right)}=\frac{-1}{3\left(x-2\right)}=\frac{-1\cdot\left(-2\right)\cdot\left(x+2\right)}{3\cdot\left(-2\right)\cdot\left(x-2\right)\cdot\left(x+2\right)}=\frac{2x+4}{-6\left(x-2\right)\left(x+2\right)}\)
\(\left(a\right)\frac{30x^3y^4}{12x^3y^4\left(x+3\right)}--\frac{7x+21}{12x^3y^4\left(x+3\right)}\)
\(\left(b\right)\frac{16x}{60x^4y^5}--\frac{55y^3}{60x^4y^5}\)
\(\left(c\right)\frac{5x-15}{2\left(x+3\right)\left(x-3\right)}--\frac{6}{2\left(x+3\right)\left(x-3\right)}\)
\(\left(d\right)\frac{6x}{3\left(x-4\right)^2}--\frac{x^2-4x}{3\left(x-4\right)^2}\)
\(\left(f\right)\frac{60}{6\left(x+2\right)}--\frac{-15}{6\left(x+2\right)}--\frac{-2}{6\left(x+2\right)}\)
\(\left(e\right)\frac{4x^2-3x+5}{x^3-1}--\frac{3x-2x^2-1}{x^3-1}--\frac{2-2x^3}{x^3-1}\)
Câu 6 :
a ) Hai phương trình không tương đương . Bởi lẽ :
\(x^2-4=0\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\) ( Nhận 2 nghiệm ) . Còn phương trình kia chỉ nhận 1 nghiệm
b ) Hai phương trình không tương đương . Bởi vì :
\(x-3=0\Rightarrow x=3\) ( một nghiệm)
\(x^2+1=0\Leftrightarrow x^2=-1\) vô lí ( vô nghiệm )
phân tích tử trc cho đỡ mất công gõ cả ps
u4-u3v+u2v2-uv3
=(u4+u2v2)-(u3v+uv3)
=u2(u2+v2)-uv(u2+v2)
=(u2-uv)(u2+v2)
=u(u-v)(u2+v2)
Thay vào ta có \(\frac{u\left(u-v\right)\left(u^2+v^2\right)}{u^2+v^2}=u\left(u-v\right)=u^2-uv\)
a. M=-1^2+2^2-3^2+4^2-...-99^2+100^2.
M=(2-1)(2+1)+(4-3)(4+3)+...+(100-99)(100+99)
M=3+7+...+199
=>2M=3+7+...+199+3+7+...+199 (198 số)
=(3+199)+(7+195)+...+(199+3) (99 cặp)
=202.99
=19998
=>M=19998:2=9999
a) m 2 − 2 m m 2 − 9 b) 3 4 ( u − 2 v ) ( u + v )