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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: \(N=\frac{2x-6}{x+3}\)
a, 3x -2 = 2x - 3
=> 3x - 2x = 2 - 3
=> x= - 1
b, là tương tự câu a
các câu sau bạn nhân phá ra mà giải nhé
a, 3x - 2 = 2x - 3
3x - 2x = -3 + 2
x = -1
b, 3 - 4u + 24 + 6u = u + 27 + 3u
-4u + 6u - u - 3u = 27 - 3 - 24
-2u = 0
u = 0 : (-2)
u = 0
c, 5 - (x - 6) = 4(3 - 2x)
5 - x + 6 = 12 - 8x
-x + 8x = 12 - 5 - 6
7x = 1
x = 1/7
d, -6(1,5 - 2x) = 3(-15 + 2x)
-9 + 12x = -45 + 6x
12x - 6x = -45 + 9
6x = -36
x = (-36) : 6
x = -6
e, 0,1 - 2(0,5 - 0,1) = 2(t - 2,5) - 0,7
0,1 - 1 + 0,2 = 2t - 5 - 0,7
-2t = -5 - 0,7 - 0,1 + 1 - 0,2
-2t = -5
t = -5/-2
t = 5/2
Câu 1:
a. \(\frac{1}{4}x^2-64\)
\(=\left(\frac{1}{2}x\right)^2-8^2\)
\(=\left(\frac{1}{2}x+8\right)\left(\frac{1}{2}x-8\right)\)
b. \(\frac{1}{27}+x^3\)
\(=\left(\frac{1}{3}\right)^3+x^3\)
\(=\left(\frac{1}{3}+x\right)\left(\frac{1}{9}-\frac{1}{3}x+x^2\right)\)
c. \(\left(a+b\right)^3-\left(a-b\right)^3\)
\(=\left(a^3+3a^2b+3ab^2+b^3\right)-\left(a^3-3a^2b+3ab^2-b^3\right)\)
\(=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b+3ab^2+b^3\)
\(=6a^2b+2b^3\)
\(=2b\left(3a^2+b^2\right)\)
a) Ta có: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(\Leftrightarrow2\cdot A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(\Leftrightarrow2\cdot A-A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(\Leftrightarrow A=1-\frac{1}{2^{100}}\)
Câu 1:
\(\dfrac{2^{35}.45^{25}.13^{22}.35^{16}}{9^{26}.65^{22}.28^{17}.25^9}\)
\(=\dfrac{2^{35}.9^{25}.5^{25}.13^{22}.7^{16}.5^{16}}{9^{26}.13^{22}.5^{22}.2^{17}.2^{17}.7^{17}.5^9.5^9}\)
Bạn rút gọn sẽ còn lại:
\(=\dfrac{2.5}{7.9}=\dfrac{10}{63}\)
Câu 4:
\(K=\left(x^2y-3\right)^2-\left(2x-y\right)^3+xy^2\left(6-x^3\right)+8x^3-6x^2y-y^3\)\(K=\left(x^2y\right)^2-2.x^2y.3+3^2-\left[\left(2x\right)^3-3.\left(2x\right)^2.y+3.2x.y^2-y^3\right]+6xy^3-x^4y^2+8x^3-6x^2y-y^3\)\(K=x^4y^2-6x^2y+9-8x^3+12x^2y-6xy^2+y^3+6xy^2-x^4y^2+8x^3-6x^2y-y^3\)\(K=9\)
a) Ta có:
\(\frac{{9{{\rm{x}}^2} + 3{\rm{x}} + 1}}{{27{{\rm{x}}^3} - 1}} = \frac{{9{{\rm{x}}^2} + 3{\rm{x}} + 1}}{{\left( {3{\rm{x}} - 1} \right)\left( {9{{\rm{x}}^2} + 3{\rm{x}} + 1} \right)}} = \frac{1}{{3{\rm{x}} - 1}}\)
\(\frac{{{x^2} - 4{\rm{x}}}}{{16 - {x^2}}} = \frac{{x\left( {x - 4} \right)}}{{\left( {4 - x} \right)\left( {4 + x} \right)}} = \frac{{ - x\left( {4 - x} \right)}}{{\left( {4 - x} \right)\left( {4 + x} \right)}} = \frac{{ - x}}{{4 + x}}\)
b) Mẫu thức chung của hai phân thức nhân được ở câu a là: \(\left( {3{\rm{x}} - 1} \right)\left( {4 + x} \right)\)
Nhân tử phụ của \(\frac{1}{{3{\rm{x}} - 1}}\) là: \(4 + x\)
Nhân tử phụ của \(\frac{{ - x}}{{4 + x}}\) là : \(3{\rm{x}} - 1\)
Khi đó:
\(\frac{1}{{3{\rm{x}} - 1}} = \frac{{4 + x}}{{\left( {3{\rm{x}} - 1} \right)\left( {4 + x} \right)}}\)
\(\frac{{ - x}}{{4 + x}} = \frac{{ - x\left( {3{\rm{x}} - 1} \right)}}{{\left( {4 + x} \right)\left( {3{\rm{x}} - 1} \right)}}\)
1)\(25x^2y^4+30xy^2z+9z^2=\left(5xy^2+3z\right)^2\)
\(\dfrac{16}{9}x^2+4xyz^2+\dfrac{9}{4}y^2z^4=\left(\dfrac{4}{3}x+\dfrac{3}{2}yz^2\right)^2\)
2)
a)\(\dfrac{9}{25}x^2+\dfrac{12}{35}xy+\dfrac{4}{49}y^2=\left(\dfrac{3}{5}x+\dfrac{2}{7}y\right)^2=\left(\dfrac{3}{5}.5+\dfrac{2}{7}.\left(-7\right)\right)^2=\left(3-2\right)^2=1\)b)\(\dfrac{25}{16}u^4v^2+\dfrac{1}{5}u^2v^3+\dfrac{4}{625}v^4\)
\(=\left(\dfrac{5}{4}u^2v+\dfrac{2}{25}v^2\right)^2=\left(\dfrac{5}{4}.\dfrac{4}{25}.\left(-5\right)+\dfrac{2}{25}.\left(-5\right)^2\right)^2\)
\(=\left(-1+2\right)^2=1\)
Thực hiện phép nhân đa thức với đa thức ở vế trái
a) VT = 3 u 2 + 9u + 27 – ( u 3 – 32 u 2 + 9u) = 27 – u 3 = VP (đpcm).
b) VT = ( t 2 – 4)( t 2 + 4) = t 4 – 16 = VP. (đpcm).