raspuns
D) 2/3-(x-1)/6+(4-3*x)/9=(x-5)/2
aducem la acelasi numitor: 2*6-(x-1)*3+(4-3x)*2=(x-5)*9
18-3x+3+8-6x=9x-45 29+45=9x+9x 18x=74 ...
e) 2*(2x-1)/3-(x-5)=x/8+(x+7)/6
aducem la acelasi numitor: 16*(2x-1)-24*(x-5)=3x+4*(x+7) 32x-16-24x+120=3x+4x+28 8x+104=7x+28 x= - 76
f)-2/3=(4x+1)/6+(x+1)/4-2
aducem la acelasi numitor: -4*2=2*(4x+1)+3*(x+1)-12*2 -8=8x+2+3x+3-24
11x-19=-8 x=11/11 x=1
7) a) m=-3 deci, (-3-1)x-2(x-3)=x-3 -4x-2x+6=x-3 -6x-x=-3-6 -7x=-9 x=9/7
b) x=2 deci, (m-1)*2-2(2-3)=2+m 2m-2+2=m+2 m=2
c) (m-1)x-2*(x-3)=x+m mx-x-2x+6=x+m mx-3x+6=x+m mx-4x=m-6 x(m-4)=m-6 pentru ca ecuatia sa nu aiba solutie reala trebuie ca m-4=0 deci m=4
8)a) a=1/2 deci: (1/2-1)(x+3)-x(2*1/2+1)=-9
aducem la acelasi numitor: -(x+3)-x*4=-9*2 -x-3-4x=-18 -5x=-18+3 -5x=-15 x=3
b) x=-3 deci: (a-1)(-3+3)-(-3)(2a+1)=-9 6a+3=-9 6a=-12 a=-2
c) (a-1)(x+3)-x(2a+1)=-9 ax+3a-x-3-2ax-x=-9 -ax-2x+3a-3=-9 -x(a+2)=-3a-9+3
-x(a+2)=-3a-6 -x(a+2)=-3(a+2) pt. ca ec. sa aiba o infinitate de solutii trebuie ca a+2=0 deci a=-2
9) b) |x|=7 deci x=-7 si x=7
c) |2x|=1 |x|=1/2 deci x=-1/2 si x=1/2
e) x-|-4|=1 x-4=1 x=5
f) 3-|x|=-5 |x|=3+5 |x|=8 x=-8 si x=8
g) |-2|+|x|=9 2+|x|=9 |x|=9-2 |x|=7 x=-7 si x=7
10) a) |x-3|=2 x-3=-2 x=3-2 x=1
si x-3=2 x=3+2 x=5
b) |2x+7|=5 2x+7=-5 2x=-5-7 x=-6
si 2x+7=5 2x=5-7 x=-1
c) |x-√3|=2√3-1 x-√3=-2√3+1 x=√3-2√3+1 x=1-√3
si x-√3=2√3-1 x=√3+2√3-1 x=3√3-1
d) 1-2*|x-1|=1/4 4-4*2*|x-1|=1 8*|x-1|=3 |x-1|=3/8
x-1=-3/8 x=1-3/8 x=5/8
si x-1=3/8 x=11/8
e) |x+3|-4=3*|x+3| (3-1)*|x+3|=-4 2*|x+3|=-4 |x+3|=-2
x+3=-(-2) x+3=2 x=-1
si x+3=-2 x=-5
f) |3x-2|+3√5=8 |3x-2|=8-3√5
3x-2=-(8-3√5) 3x=2-8+3√5 3x=3√5-6 x=√5-2
si 3x-2=8-3√5 3x=2+8-3√5 3x=10-3√5 x=10/3-√5
11) a) (x-1)²=9 x²-2x+1=9 x²-2x-8=0 discriminantul: Δ=(-2)²-4*1*(-8)=4+32=36
radacinile x1=(2-√36)/(2*1)=-4/2=-2
x2=(2+√36)/(2*1)=8/2=4
b) 3-(x-5)²=1 3-(x²-2x*5+25)=1 3-x²+10x-25=1 x²-10x+23=0
Δ=(-10)²-4*1*23=100-92=8 solutiile sunt x1=(10-√8)/(2*1)=(10-2√2)/2=5-√2
si x2=(10+√8)/2=5+√2
c) (3-x)(3+x)-6=2 3²-x²-6=2 x²-1=0 Δ=-4*1*(-1)=4
x1=(-√4)/2=-1 si x2=√4/2=1
d) 5-(x-2)(x+2)=3 5-(x²-2²)=3 5-x²+4=3 x²-6=0 Δ=-4*(-6)=24
x1=-√24/2=-2√6/2=-√6 si x2=√24/2=√6
e) x²-2x+1=7 x²-2x-6=0 Δ=(-2)²-4*(-6)=4+24=28
x1=(2-√28)/2=(2-2√7)/2=1-√7 si x2=(2+√28)/2=1+√7
f) 4x²-4x+1=5 4x²-4x-4=0 x²-x-1=0 Δ=(-1)²-4*(-1)=1+4=5
x1=(1-√5)/2 si x2=(1+√5)/2
12) a) √x=25 conditia: x mai mare ca 0, ridicam la a 2-a: x=625
b) √(3x-2)=5 conditia: 3x-2≥0, x≥2/3
ridicam la putere ecuatia: 3x-2=25 3x=27 x=9≥2/3
c) 6-√(2*x)=2 conditia: 2x mai mare ca 0
√(2x)=6-2 2x=16 x=8
d) √(2x-3)=25 conditia: 2x-3≥0 x≥3/2
ridicam la putere: 2x-3=625 2x=628 x=314≥3/2
e) √x²+1=10 conditia: x² mai mare ca 0
√x²-9=0 x-9=0 x=9
f) 3+√(x+4)=2 trebuie ca x+4≥2 deci x≥-2
√(x+4)=2-3 √(x+4)=-1 cum -1 e mai mic ca 0 ecuatia nu are solutii
m am chinuit
Explicație pas cu pas:
sper ca e okk
