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\(\left(5\cdot\left(x^2-3x+1\right)+x\cdot\left(1-5x\right)\right)-\left(x-2\right)=0\)
\(7-15x=0\)
\(-15x=-7\)
\(x=\frac{7}{15}=0.467\)
\(b,\)câu b dài quá nên mik lười, vậy mik ghi kết quả thôi nhé
\(x=\frac{2}{19}=0.105\)
\(c,\)câu c cũng vậy mik ghi kết quả thôi nhé bn
\(x=-\frac{6}{11}=-0.545\)
1
\(-3x\left(x-5\right)+5\left(x-1\right)+3x^2=4-x\)
=> \(-3x^2+15x+5x-5+3x^2=4-x\)
=> \(20x-5=4-x\)
=> \(21x=9\)
=> \(x=\dfrac{3}{7}\)
Vậy x = \(\dfrac{3}{7}\)
2,
\(7x\left(x-2\right)-5\left(x-1\right)=21x^2-14x^2+3\)
=> \(7x^2-14x-5x+5=7x^2+3\)
=> \(-14x-5x+5=3\)
=> \(-19x=-2\)
=> \(x=\dfrac{2}{19}\)
Vậy \(x=\dfrac{2}{19}\)
3,
\(3\left(5x-1\right)-x\left(x-2\right)+x^2-13x=7\)
=> \(15x-3-x^2+2x+x^2-13x=7\)
=> \(4x-3=7\)
=> 4x = 10
=> x = \(\dfrac{5}{2}\)
Vậy x = \(\dfrac{5}{2}\)
4,
\(\dfrac{1}{5}x\left(10x-15\right)-2x\left(x-5\right)=12\)
=> \(2x^2-3x-2x^2+10x=12\)
=> 7x = 12
=> x = \(\dfrac{12}{7}\)
Vậy x = \(\dfrac{12}{7}\)
Kkk
a) \(7x\left(x-2\right)-2\left(x-1\right)=21x^2-14x^2+3\)
\(\Leftrightarrow7x^2-14x-2x+2=7x^2+3\)
\(\Leftrightarrow7x^2-16x+2-7x^2-3=0\)
\(\Leftrightarrow-16x=1\)
\(\Leftrightarrow x=-\frac{1}{16}\)
b) \(3\left(5x-1\right).x\left(x-2\right)+x^2-13x=7\)
\(\Leftrightarrow\left(15x-3\right)\left(x^2-2x\right)+x^2-13x=7\)
\(\Leftrightarrow15x^3-30x^2-3x^2+6x+x^2-13x=7\)
\(\Leftrightarrow15x^3-32x^2-7x-7=0\)
Phân tích đa thức thành nhân tử, ra nghiệm vô tỉ:)
c) \(\frac{1}{5}x\left(10x-5\right)-2x\left(x-5\right)=15\)
\(\Leftrightarrow2x^2-x-2x^2+10x=15\)
\(\Leftrightarrow9x=15\)
\(\Leftrightarrow x=\frac{5}{3}\)
1 ) Thực hiện phép tính :
a ) \(-\frac{1}{3}xz\left(-9xy+15yz\right)+3x^2\left(2yz^2-yz\right)\)
\(=3x^2yz-5xyz^2+6x^2yz^2-3x^2yz\)
\(=-5xyz^2+6x^2yz^2\)
b ) \(\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\)
\(=x^3-5x^2-x-2x^2+10x-2-x^3-11x\)
\(=-7x^2-2x-2-x^3\)
c ) \(\left(x^3+5x^2-2x+1\right)\left(x-7\right)\)
\(=x^4+5x^3-2x^2+x-7x^3-35x^2+14x-7\)
\(=x^4-2x^3-37x^2+15x-7\)
d ) \(\left(2x^2-3xy+y^2\right)\left(x+y\right)\)
\(=2x^3-3x^2y+xy^2+2x^2y-3xy^2+y^3\)
\(=2x^3-x^2y-2xy^2+y^3\)
e ) \(\left[\left(x^2-2xy+2y^2\right)\left(x+2y\right)-\left(x^2-4y^2\right)\left(x-y\right)\right]2xy\)
( để xem lại )
2 Tìm x
a ) \(6x\left(5x+3\right)+3x\left(1-10x\right)=7\)
\(\Leftrightarrow30x^2+18x+3x-30x^2=7\)
\(\Leftrightarrow21x=7\)
\(\Leftrightarrow x=3\)
b ) Sai đề
c ) \(\left(x+1\right)\left(x+2\right)\left(x+5\right)-x^2\left(x+8\right)=27\)
( Để xem lại )
mình chép đúng theo đề cô cho mà sao lại sai được ,hay cô cho sai đề
1/
a. \(3x\left(5x^2-2x-1\right)\)
\(=15x^3-6x^2-3x\)
b. \(\left(x^2-2xy+3\right)\left(-xy\right)\)
\(=-x^3y+2x^2y^2-3xy\)
c. \(\left(2x^2-3xy+y^2\right)\left(x+y\right)\)
\(=2x^3-3x^2y+xy^2+2x^2y-3xy^2+y^3\)
\(=2x^3-x^2y-2xy^2\)
a) thiếu đề
b) \(\left(3x-3\right)\left(5-21x\right)+\left(7x+4\right)\left(9x-5\right)=44\)
\(15x-63x^2-15+63x+63x^2-35x+36x-20=44\)
\(79x-35=40\)
\(79x=75\)
\(x=\frac{75}{79}\)
a) Ta có: \(-3x^2\left(2x^2-\frac{1}{3}x+2\right)\)
\(=-6x^4+x^3-6x^2\)
b) Ta có: \(2xy^2\left(x-3y+xy\right)\)
\(=2x^2y^2-6xy^3+2x^2y^3\)
c) Ta có: \(\left(5x^2-4x\right)\left(x-2\right)\)
\(=5x^3-10x^2-4x^2+8x\)
\(=5x^3-14x^2+8x\)
d) Ta có: \(-\left(2-x\right)\left(2x+3\right)\)
\(=\left(x-2\right)\left(2x+3\right)\)
\(=2x^2+3x-4x-6\)
\(=2x^2-x-6\)
e) Ta có: \(\left(3x^3-2x^2+x\right):\left(-2x\right)\)
\(=\frac{-3}{2}x^2+x-\frac{1}{2}\)
f) Ta có: \(\left(15x^2y^2-21x^3y+2x^2y\right):\left(3x^2y\right)\)
\(=5y-7x+\frac{2}{3}\)
g) 
\(1,5\left(x^2-3x+1\right)+x\left(1-5x\right)=x-2\)
\(\Leftrightarrow5x^2-15x+5+x-5x^2-x+2=0\)
\(\Leftrightarrow-15x^2+7=0\)
\(\Leftrightarrow x^2=\frac{7}{15}\)
\(\Leftrightarrow x=\overset{+}{-}\sqrt{\frac{7}{15}}\)
\(2,7x\left(x-2\right)-5\left(x-1\right)=21x^2-14x^2+3\)
\(\Leftrightarrow7x^2-14x-5x+5-7x-3=0\)
\(\Leftrightarrow7x^2-26x+2=0\)
\(\Leftrightarrow7\left(x^2-\frac{26}{7}x+\frac{169}{49}\right)-\frac{155}{7}=0\)
\(\Leftrightarrow7\left(x-\frac{13}{7}\right)^2-\frac{155}{7}=0\)
\(\Leftrightarrow\left(\sqrt{7}x-\frac{13}{\sqrt{7}}-\sqrt{\frac{155}{7}}\right)\left(\sqrt{7}x-\frac{13}{\sqrt{7}}+\sqrt{\frac{155}{7}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{155}+13}{7}\\x=\frac{-\sqrt{155}+13}{7}\end{matrix}\right.\)