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\(\frac{x-4}{-5}=\frac{1-2x}{3}\)
Nhân cả 2 vế với 15 ,ta được:
\(\frac{15.\left(x-4\right)}{-5}=\frac{15.\left(1-2x\right)}{3}\)
\(\Leftrightarrow\left(-3\right).\left(x-4\right)=5.\left(1-2x\right)\)
\(\Leftrightarrow-3x+12=5-10x\)
\(\Leftrightarrow-3x+10x=5-12\)
\(\Leftrightarrow7x=-7\)
\(\Leftrightarrow x=-1\)
Vậy x=-1
(𝑥−4)/−5=(1−2𝑥)/3
−15⋅𝑥−4−5=−15⋅−2𝑥+13
−15⋅𝑥−4−5=−15⋅−2𝑥+13
3(𝑥−4)=−5(−2𝑥+1)
3(x-4)=-5(-2x+1)
3(𝑥−4)=−5(−2𝑥+1)
3𝑥−12=−5(−2𝑥+1)
3𝑥−12=−5(−2𝑥+1)
3𝑥−12=10𝑥−5
𝑥 = -1
\(\left(\frac{3}{5}\right)^2=\frac{9}{25}\)
\(n=5.2=10\)
\(\text{chúc e hok tốt nhé}\)
( x -1 ) ( x+1 )=0
=> x +1 = 0 hoặc x-1 = 0
* Trường hợp 1 : Nếu x - 1 = 0 thì : x = 0 +1 = 1
* Trường hợp 2 : Nếu x + 1 = 0 thì : x = 0 - 1 = -1
Vậy x = 1 hoặc x = -1
k Đúng cho mình nha , thanks trước
\(\frac{3}{5}-\frac{1}{7}+\frac{2}{5}-\frac{6}{7}-\frac{1}{4}\)
\(=\left(\frac{3}{5}+\frac{2}{5}\right)-\left(\frac{1}{7}+\frac{6}{7}\right)-\frac{1}{4}\)
\(=\frac{5}{5}-\frac{7}{7}-\frac{1}{4}\)
\(=1-1-\frac{1}{4}\)
\(=0-\frac{1}{4}\)
\(=-\frac{1}{4}\)
\(\frac{3}{5}-\frac{1}{7}+\frac{2}{5}-\frac{6}{7}-\frac{1}{4}\)
\(=\left(\frac{3}{5}+\frac{2}{5}\right)-\left(\frac{1}{7}+\frac{6}{7}\right)-\frac{1}{4}\)
\(=1-1-\frac{1}{4}\)
\(=-\frac{1}{4}\)
a) \(-\left(-a+c-d\right)-\left(c-a+d\right)\)
\(=a-c+d-c+a-d\)
\(=2a-2c\)
b) \(-\left(a+b-c+d\right)+\left(a-b-c-d\right)\)
\(=-a-b+c-d+a-b-c-d\)
\(=-2b-2d\)
c) \(a\left(b-c-d\right)-a\left(b+c-d\right)\)
\(=ab-ac-ad-ab-ac+ad\)
\(=-2ac\)
d) \(\left(a-b\right)+\left(c-d\right)+\left(a+c\right)-\left(b+d\right)\)
\(=a-b+c-d+a+c-b-d\)
\(=2a-2b+2c-2d\)
\(-\frac{1}{7}+\frac{5}{3}+\frac{5}{4}+\frac{1}{3}-\frac{3}{2}\)
\(=\left(-\frac{1}{7}+\frac{5}{3}-\frac{3}{2}\right)+\left(\frac{5}{3}+\frac{1}{3}\right)\)
\(=\frac{-6}{42}+\frac{70}{42}-\frac{63}{42}+\frac{6}{3}\)
\(=\frac{-6+70-63}{42}+2\)
\(=\frac{1}{42}+\frac{84}{42}\)
\(=\frac{85}{42}\)
Vì \(\left|x^2+2x\right|\ge0;\left|y^2-9\right|\ge0\)
Dấu ''='' xảy ra <=> \(x^2+2x=0\Leftrightarrow x\left(x+2\right)=0\Leftrightarrow x=0;x=-2\)
\(y^2-9=0\Leftrightarrow\left(y-3\right)\left(y+3\right)=0\Leftrightarrow y=\pm3\)
Ta có :
∣∣x2+2x∣∣+∣∣y2−9∣∣=0|x2+2x|+|y2-9|=0
Do {|x2+2x|≥0|y2−9|≥0{|x2+2x|≥0|y2−9|≥0
→∣∣x2+2x∣∣+∣∣y2−9∣∣≥0→|x2+2x|+|y2-9|≥0
Mà ∣∣x2+2x∣∣+∣∣y2−9∣∣=0|x2+2x|+|y2-9|=0
→→ {|x2+2x|=0|y2−9|=0{|x2+2x|=0|y2−9|=0
→→ {x2+2x=0y2−9=0{x2+2x=0y2−9=0
→→ {x(x+2)=0y2=9{x(x+2)=0y2=9
→→ ⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩[x=0x+2=0[y=3y=−3{[x=0x+2=0[y=3y=−3
→→ ⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩[x=0x=−2[y=3y=−3{[x=0x=−2[y=3y=−3
Vậy x,y∈{0;3};{0;−3};{−2;3};{−2;−3}x,y∈{0;3};{0;-3};{-2;3};{-2;-3}
a: Bậc là 3
Giải ra đi