-2x+3{12-2[3x-20-2x-4x]+1}=45
-2x+3[12-2(-3x-20)+1]=45
-2x+3(12+6x+40+1)=45
16x+159=45
16x=-114
x=-7,125
-2x+3{12-2[3x-20-2x-4x]+1}=45
-2x+3[12-2(-3x-20)+1]=45
-2x+3(12+6x+40+1)=45
16x+159=45
16x=-114
x=-7,125
a) căn(x²+12)+5=3x+căn(x²+5)
b) 9(căn(4x+1)-căn(3x-2))=x+3
c) căn(2x+4)-2 căn(2x-1)=6x-4/căn(x²+4)
d) x²+9x+20=2 căn(3x+10)
Giai phuong trinh
1/ \(\sqrt{x^2+4x+5}+\sqrt{x^2-6x+13}=3\)
2/ \(\sqrt{3x^2-18x+28}+\sqrt{4x^2-24x+45}=6x-x^2-5\)
3/ \(\sqrt{2x^2-4x+27}+\sqrt{3x^2-6x+12}=4x^2+8x+4\)
4/ \(\sqrt{x^2+x+7}+\sqrt{x^2+x+2}=\sqrt{3x^2+3x+19}\)
5/ \(\left(x+2\right)\left(x+3\right)-\sqrt{x^2+5x+1}=9\)
6/ \(\left(x+4\right)\left(x+1\right)-3\sqrt{x^2+5x+2}=6\)
7/ \(\sqrt{2x^2+3x+5}+\sqrt{2x^2-3x+5}=3\sqrt{x}\)
a)\(\sqrt{1-x}\left(x-3x^2\right)=x^3-3x^2+2x+6\)
b)\(x^2+x+12\sqrt{x+1}=36\)
c)\(3x-1+\frac{x-1}{4x}=\sqrt{3x+1}\)
d)\(\sqrt{x^2+12}-3x=\sqrt{x^2+5}-5\)
e)\(4x^2+12+\sqrt{x-1}=4\left(x\sqrt{5x-1}+\sqrt{9-5x}\right)\)
f)\(4x^3-25x^2+43x+x\sqrt{3x-2}=22+\sqrt{3x-2}\)
g)\(2\left(x+1\right)\sqrt{x}+\sqrt{3\left(2x^3+5x^2+4x+1\right)}=5x^3-3x^2+8\)
h)\(\sqrt{x^2+12}-\sqrt{x^2+5}=3x-5\)
i)\(\sqrt{1-3x}-\sqrt[3]{3x-1}=\left|6x-2\right|\)
k)\(\sqrt{2x^3+3x^2-1}=2x^2+2x-x^3-1\)
l)\(\sqrt{x^2+x-2}+x^2=\sqrt{2\left(x-1\right)}+1\)
Giải phương trình
1, \(x^2+\left(3-\sqrt{x^2+2}\right)x=1+2\sqrt{x^2+2}\)
2, \(10x^2+3x+1=\sqrt{x^2+3}\left(1+6x\right)\)
3, \(\sqrt{2x-3}+\sqrt{5-2x}=3x^2-12x+14\)
4, \(x^2+2x+15=6\sqrt{4x+5}\)
5, \(\sqrt{2x^2+5x+12}-x=5-\sqrt{2x^2+3x+2}\)
a ) \(x^4+2x^3-4x^2-2x+ 1=0\)
b)\(x^4-3x^3+4x^2-3x+1=0\)
giải các phương trình sau :
a ) \(7\sqrt{4x^2+5x-1}-14\sqrt{x^2-3x+3}=17x-13\)
b ) \(\sqrt{2x^2+5x+12}+\sqrt{2x^2+3x+2}=x+5\)
\(\frac{12}{x^{^2}+4x+2}-\frac{3x}{x^{^2}+2x+2}=1\)
Giải phương trình:
1, \(x^2+2x\sqrt{x-\dfrac{1}{x}}=3x+1\)
2, \(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{16x-4x^2-15}\)
3, \(7\sqrt{3x-7}+\left(4x-7\right)\sqrt{7-x}=32\)
giải hpt
\(\left\{{}\begin{matrix}\frac{4}{2x+y}+\frac{1}{3x-y}=2\\4x+12=7\left(2x+y\right)\left(3x-y\right)\end{matrix}\right.\)