tìm x biết
6x+5/5x-3=38-6x/21-5x
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\(\dfrac{6x+5}{5x-3}=\dfrac{38-6x}{21-5x}\left(đk:x\ne\dfrac{5}{3},x\ne\dfrac{21}{5}\right)\)
\(\Rightarrow\left(6x+5\right)\left(21-5x\right)=\left(5x-3\right)\left(38-6x\right)\)
\(\Rightarrow-30x^2+101x+105=-30x^2+208x-114\)
\(\Rightarrow107x=219\Rightarrow x=\dfrac{219}{107}\left(tm\right)\)
\(\dfrac{6x+5}{5x-3}=\dfrac{6x-38}{5x-21}\)
\(\Leftrightarrow30x^2-126x+25x-105=30x^2-18x-190x+114\)
\(\Leftrightarrow-101x-105=-208x+114\)
\(\Leftrightarrow107x=219\)
hay \(x=\dfrac{219}{107}\)
Ta có: \(VT=\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}\)
\(=\sqrt{3x^2+6x+3+4}+\sqrt{5x^2+10x+5+16}\)
\(=\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+16}\ge2+4=6\)
Ta có: \(VP=5-x^2-2x\)
\(=-\left(x^2+2x+1\right)+6\)
\(=-\left(x+1\right)^2+6\le6\)
VP=VT khi x+1=0
hay x=-1
Vậy: x=-1
Tìm x,y,z biết 6x 4z 5 2y 5x 6 5z 6y 4và 3x 2y 5z 96 tìm x,y,z biết 6x 4z 5 2y 5x 6 5z 6y 4 và 3x 2y
\((x-2)^3-(x+5)(x^2-5x+25)+6x^2=11\\\Leftrightarrow (x-2)^3-(x+5)(x^2-5.x+5^2)+6x^2=11 \\\Leftrightarrow x^3-6x^2+12x-8 -(x^3+5^3)+6x^2-11=0 \\\Leftrightarrow 12x-144=0 \\\Leftrightarrow x=12\)
Vậy \(x=12\).
(x−2)3−(x+5)(x2−5x+25)+6x2=11
=>(x−2)3−(x+5)(x2−5.x+52)+6x2=11
=>x3−6x2+12x−8−(x3+53)+6x2−11=0
=>12x−144=0
=>x=12(x−2)3−(x+5)(x2−5x+25)+6x2=11
=>(x−2)3−(x+5)(x2−5.x+52)+6x2=11
=>x3−6x2+12x−8−(x3+53)+6x2−11=0
=>12x−144=0
=>x=12
Vậy x=12x=12.
cho tôi đúng đi
\(1,\\ a,=-35x^5y^4z\\ b,=6x^2-30x-6x^2-3x=-33x\\ c,=x^3-9x^2-2x^2+18x-x+9=x^3-11x^2+17x+9\\ 2,\\ A\left(x\right)+B\left(x\right)=10-2x+4x^3-5x^2-10x^3-5x+6x^2-20\\ =-6x^3+x^2-7x-10\\ A\left(x\right)-B\left(x\right)=10-2x+4x^3-5x^2+10x^3+5x-6x^2+20\\ =14x^3-11x^2+3x+30\\ 3,\\ a,M\left(x\right)=5x+20=0\\ \Leftrightarrow x=-4\\ b,N\left(x\right)=100x^2-49=0\\ \Leftrightarrow\left(10x-7\right)\left(10x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{10}\\x=-\dfrac{7}{10}\end{matrix}\right.\\ c,P\left(x\right)=3x-15=0\\ \Leftrightarrow x=5\)
Bài 1;
a)\(5x^3yz.\left(-7x^2y^3\right)=-35.x^5y^4z\)
b)\(6x\left(x-5\right)-x\left(6x+3\right)=6x^2-30x-6x^2-3x=-33x\)
c) \(\left(x-9\right)\left(x^2-2x-1\right)=x^3-2x^2-x-9x^2+18x+9=x^3-11x^2+17x+9\)
Bài giải
a, \(-\left(-x\right)-\left(-9\right)=-\left(3-6-9+3\right)\)
\(x+9=-3+6+9-3\)
\(x+9=15\)
\(x=15-9\)
\(x=6\)
b, \(17-x=7-6x\)
\(6x-x=7-17\)
\(5x=-10\)
\(x=-10\text{ : }5\)
\(x=-2\)
c, \(5x-7=-21-2x\)
\(5x+2x=-21+7\)
\(7x=-14\)
\(x=-14\text{ : }7\)
\(x=-2\)
d, \(7\left(x-3\right)-5\left(3-x\right)=11x-5\)
\(7x-21-15+5x=11x-5\)
\(7x+5x-11x=21+15-5\)
\(x=31\)
a) \(5x^3-125=0\)
\(\Leftrightarrow5x^3=125\)
\(\Leftrightarrow x^3=25\)
\(\Leftrightarrow x^3=25\)
\(\Leftrightarrow x=\sqrt[3]{25}\)
c) \(6x+13x+5=0\)
\(\Leftrightarrow19x+5=0\)
\(\Leftrightarrow19x=-5\)
\(\Leftrightarrow x=\dfrac{-5}{19}\)
a. (5x-1)2 - (5x-4) (5x-4) +7
= (5x-1)2 - (5x-4)2 + 7
=[(5x-1)+(5x-4)] [(5x-1)-(5x-4)] +7 ( đoạn này bỏ cx đc)
=(10x-5) .3+7
=30x-15+7
=30x-8