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giải
a)4x^2-20x-(4x^2+3x-4x-3)=5
4x^2-20x-4x^2-3x+4x+3=5
-19x+3=5
-19x=5-3
-189x=2
x=-2/19
mik giải luôn đó chứ ko viết đầu bài đâu
c)
2x(x-3)-2(x^2-4)=4
2x^2-6x-2x^2+8=4
-6x+8=44
-6x=4-8
-6x=-4
x=2/3
a) Ta có: \(-5x^2+3x=0\)
\(\Leftrightarrow x\left(-5x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-5x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{3}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\frac{3}{5}\right\}\)
b) Ta có: \(1+\frac{x-1}{3}=\frac{2x+1}{6}-2\)
\(\Leftrightarrow1+\frac{x-1}{3}-\frac{2x+1}{6}+2=0\)
\(\Leftrightarrow3+\frac{x-1}{3}-\frac{2x+1}{6}=0\)
\(\Leftrightarrow\frac{18}{6}+\frac{2\left(x-1\right)}{6}-\frac{2x+1}{6}=0\)
\(\Leftrightarrow18+2x-2-2x-2=0\)
\(\Leftrightarrow14=0\)(vô lý)
Vậy: x∈∅
c) Ta có: 2-x=3(x+1)
⇔2-x=3x+3
⇔2-x-3x-3=0
⇔-4x-1=0
⇔-4x=1
hay \(x=\frac{-1}{4}\)
Vậy: \(x=\frac{-1}{4}\)
d) Ta có: 4x+7(x-2)=-9x+5
⇔4x+7x-14+9x-5=0
⇔20x-19=0
⇔20x=19
hay \(x=\frac{19}{20}\)
Vậy: \(x=\frac{19}{20}\)
e) Ta có: -4(x+3)=5(2x-9)
⇔-4x-12=10x-45
⇔-4x-12-10x+45=0
⇔-14x+33=0
⇔-14x=-33
hay \(x=\frac{33}{14}\)
Vậy: \(x=\frac{33}{14}\)
f) Ta có: \(\frac{2x-1}{3}-\frac{5x+2}{4}=2x\)
\(\Leftrightarrow\frac{4\left(2x-1\right)}{12}-\frac{3\left(5x+2\right)}{12}=\frac{24x}{12}\)
\(\Leftrightarrow4\left(2x-1\right)-3\left(5x+2\right)-24x=0\)
\(\Leftrightarrow8x-4-15x-6-24x=0\)
\(\Leftrightarrow-31x-10=0\)
\(\Leftrightarrow-31x=10\)
hay \(x=\frac{-10}{31}\)
Vậy: \(x=\frac{-10}{31}\)
Dài dữ trời :V Về sau gửi từng bài một thôi, nhìn hoa mắt quá @@
B1: Phân tích thành nhân tử:
a) \(6x^2+9x=3x\left(2x+3\right)\)
b) \(4x^2+8x=4x\left(x+2\right)\)
c) \(5x^2+10x=5x\left(x+2\right)\)
d) \(2x^2-8x=2x\left(x-4\right)\)
e) \(5x-15y=5\left(x-3y\right)\)
f) \(x\left(x^2-1\right)+3\left(x^2-1\right)=\left(x^2-1\right)\left(x+3\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+3\right)\)
g) \(x^2-2x+1-4y^2=\left(x-1\right)^2-4y^2\)
\(=\left(x-1-2y\right)\left(x-1+2y\right)\)
h) \(x^2-100=\left(x-10\right)\left(x+10\right)\)
i) \(9x^2-18x+9=\left(3x-3\right)^2\)
k) \(x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)
l) \(x^2+6xy^2+9y^4=\left(x+3y\right)^2\)
m) \(4xy-4x^2-y^2=-\left(4x^2-4xy+y^2\right)\)
\(=-\left(2x-y\right)^2\)
n) \(\left(x-15\right)^2-16=\left(x-15-16\right)\left(x-15+16\right)\)
\(=\left(x-31\right)\left(x+1\right)\)
o) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3+x\right)\)
\(=\left(2+x\right)\left(8+x\right)\)
p) \(\left(7x-4\right)^2-\left(2x+1\right)^2\)
\(=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)\)
\(=\left(5x-5\right)\left(9x-3\right)\)
Bài 1 :
a ) \(6x^2+9x=3x\left(x+3\right)\)
b ) \(4x^2+8x=4x\left(x+2\right)\)
c ) \(5x^2+10x=5x\left(x+2\right)\)
d ) \(2x^2-8x=2x\left(x-4\right)\)
e ) \(5x-15y=5\left(x-3y\right)\)
f ) \(x\left(x^2-1\right)+3\left(x^2-1\right)=\left(x^2-1\right)\left(x+3\right)\)
g ) \(x^2-2x+1-4y^2=\left(x-1\right)^2-\left(2y\right)^2=\left(x-1-2y\right)\left(x-1+2y\right)\)
h ) \(x^2-100=x^2-10^2=\left(x-10\right)\left(x+10\right)\)
i ) \(9x^2-18x+9=\left(3x-3\right)^2\)
k ) \(x^3-8=\left(x-2\right)\left(x^2+2x+2^2\right)\)
l ) \(x^2+6xy^2+9y^4=\left(x+3y^2\right)^2\)
m ) \(4xy-4x^2-y^2=-\left(2x-y\right)^2\)
n ) \(\left(x-15\right)^2=x^2-30x+15^2\)
o ) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3-x\right)=\left(2+x\right)\left(8-x\right)\)
p ) \(\left(7x-4\right)^2-\left(2x+1\right)^2=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)=\left(5x-5\right)\left(9x-3\right)\)
Bài 2 :
a ) \(3x^3-6x^2+3x^2y-6xy=3x\left(x^2-2x+xy-2y\right)\)
b ) \(x^2-2x+xy-2y=x\left(x-2\right)+y\left(x-2\right)=\left(x-2\right)\left(x+y\right)\)
c ) \(2x+x^2-2y-2xy=......................\)
d ) \(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
e ) \(x^2+y^2-2xy-4=\left(x-y\right)^2-2^2=\left(x-y-2\right)\left(x-y+2\right)\)
f )\(2xy-x^2-y^2+9=-\left(x-y\right)^2+9=3^2-\left(x-y\right)^2=\left(3-x+y\right)\left(3+x-y\right)\)
Lần sau đăng thì chia thành nhiều câu hỏi nhé
\(16^2-9.\left(x+1\right)^2=0\)
\(16^2-\text{ }\left[3.\left(x+1\right)\right]^2=0\)
\(\left[16-3.\left(x+1\right)\right].\left[16+3\left(x+1\right)\right]=0\)
\(\left[16-3x-3\right]\left[16+3x+3\right]=0\)
\(\left[13-3x\right].\left[19+3x\right]=0\)
\(\Rightarrow\orbr{\begin{cases}13-3x=0\\19+3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=13\\3x=-19\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{13}{3}\\x=-\frac{19}{3}\end{cases}}}\)
KL:..............................
a) \(\left(2x+1\right)\left(3x-2\right)=\left(2x+1\right)\left(5x-8\right)\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(3x-2\right)-\left(2x+1\right)\left(5x-8\right)=0\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(3x-2-5x+8\right)=0\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(6-2x\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}2x+1=0\\6-2x=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-0,5\\x=3\end{cases}}\)
Vậy...
b) \(ĐKXĐ:\) \(x\ne-2;\) \(x\ne4\)
\(\frac{3}{x+2}+\frac{2}{x-4}=0\)
\(\Leftrightarrow\)\(\frac{3\left(x-4\right)}{\left(x+2\right)\left(x-4\right)}+\frac{2\left(x+2\right)}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\)\(\frac{3x-12+2x+4}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\)\(\frac{5x-8}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Rightarrow\)\(5x-8=0\)
\(\Leftrightarrow\)\(x=\frac{8}{5}\) (T/m đkxđ)
Vậy...
c) \(x^3+4x^2+4x+3=0\)
\(\Leftrightarrow\)\(x^3+3x^2+x^2+3x+x+3=0\)
\(\Leftrightarrow\)\(x^2\left(x+3\right)+x\left(x+3\right)+\left(x+3\right)=0\)
\(\Leftrightarrow\)\(\left(x+3\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\)\(x+3=0\) (do \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\) \(\forall x\))
\(\Leftrightarrow\)\(x=-3\)
Vậy...
a) \(2x\left(2x+5\right)-4x\left(x-3\right)=7\)
\(4x^2+10x-4x^2+12x=7\)
\(22x=7\Rightarrow x=0,31\)
b) \(\left(x+2\right)\left(x-2\right)-\left(x+1\right)^2=2\)
\(\left(x^2-4\right)-\left(x^2+2x+1\right)=2\)
\(x^2-4-x^2-2x-1=2\)
\(-2x=7\Rightarrow x=-3,5\)
c) \(\left(x+2\right)\left(x-1\right)-\left(x+3\right)\left(x-2\right)=0\)
\(x^2-x+2x-2-x^2+2x+3x-6=0\)
\(6x=8\Rightarrow x=1,3\)