a)(a+2)2\(\left|3b-4\right|\)=0
vì (a+2)2 \(\ge\) =0 \(\forall\) a
\(\left|3b-4\right|\)\(\ge0\forall b\)
=>(a+2)2=0 và \(\left[3b-4\right]\) =0
=> a-2=0 và 3b-4=0
=>a=2 và b=\(\dfrac{4}{3}\)
Nếu \(a\left(y+z\right)=b\left(z+x\right)=c\left(x+y\right)\left(1\right)\)
Trong đó a, b, c là các số khác nhau và khác 0 thì: \(\dfrac{y-z}{a\left(b-c\right)}=\dfrac{z-x}{b\left(c-a\right)}=\dfrac{x-y}{c\left(a-b\right)}\)(*)
Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\) . Chứng minh :
a) \(\dfrac{3a+5b}{2a-7b}=\dfrac{3c+5d}{2c-7d}\)
b) \(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{ab}{cd}\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}.CMR\)
a, \(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)
b, \(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)
c, \(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{a^2+b^2}{c^2+d^2}\)
Tìm x biết :
a) \(\frac{-5}{7}\)-\((\frac{1}{2}-x)\)=\(\frac{-11}{4}\)
b)\((2x-1)^2-5=20\)
c)\(\left(\frac{3}{4}x+3\right)-\left(\frac{2}{3}x-4\right)-\left(\frac{1}{6}x+1\right)=\left(\frac{1}{3}x+4\right)-\left(\frac{1}{3}x-3\right)\)
d)\(\frac{x-6}{4}=\frac{4}{x-6}\)
e)\(\frac{2x-1}{3}=\frac{-27}{1-2x}\)
f)\(\left(x-1\right)^{x+2}=\left(x-1\right)^2\)
Tìm x biết :
a) \(\frac{-5}{7}\)-\((\frac{1}{2}-x)\)=\(\frac{-11}{4}\)
b)\((2x-1)^2-5=20\)
c)\(\left(\frac{3}{4}x+3\right)-\left(\frac{2}{3}x-4\right)-\left(\frac{1}{6}x+1\right)=\left(\frac{1}{3}x+4\right)-\left(\frac{1}{3}x-3\right)\)
d)\(\frac{x-6}{4}=\frac{4}{x-6}\)
e)\(\frac{2x-1}{3}=\frac{-27}{1-2x}\)
f)\(\left(x-1\right)^{x+2}=\left(x-1\right)^2\)
CM
\(\dfrac{\left(a-b\right)^3}{\left(c-d\right)^3}=\dfrac{3a^2+2b^2}{3c^2+2b^2}\)
Bài 1: Cho \(\frac{a}{b}=\frac{c}{d}\) .CM:
a) \(\frac{a^2}{a^2+b^2}=\frac{c^2}{c^2+d^2}\) b) \(\left(\frac{a+c}{b+d}\right)^2=\frac{a^2+c^2}{b^2+d^2}\)
Bài 2: Cho 3 số a,b,c\(\ne\)0, sao cho a\(^2\)=bc. CM:
a) \(\frac{a^2+c^2}{b^2+a^2}=\frac{c}{b}\) b)\(\left(\frac{c+2019a}{a+2019b}\right)^2=\frac{c}{b}\)
Bài 4: Cho a,b,c,d khác 0 sao cho b2=ac, c2=bd.CM: \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\)
1.Tính
a) 5 - \(\left(-\dfrac{15}{11}\right)^0+\left(\dfrac{1}{3}\right)^2:3\)
b) \(2^3+3.\left(\dfrac{1}{2}\right)^0+[\left(-2\right)^2:\dfrac{1}{2}].8\)
2.Tính giá trị biểu thức
a) \(\dfrac{6^3+3.6^2+3^3}{-13}\)
GIÚP MÌNH VỚI NHA CHIỀU NAY MÌNH PHẢI NỘP RỒI
Cho a, b, c, d là 4 số khác 0 thỏa mãn \(b^2\) = ac; \(c^2\) = bd và \(b^3+c^3+d^3\ne0\)
Chứng minh rằng: \(\dfrac{a}{d}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\)
