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\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
\(\Rightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)
\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)
\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)
\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)
\(\Rightarrow x+10=0\) vì \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)
\(\Rightarrow x=-10\)
a. \(\left(\frac{1}{3}x\right):\frac{2}{3}=1\frac{3}{4}:\frac{2}{5}\)
\(\left(\frac{1}{3}x\right):\frac{2}{3}=\frac{35}{8}\)
\(\frac{1}{3}x=\frac{35}{8}.\frac{2}{3}=\frac{35}{12}\)
\(x=\frac{35}{12}.3=\frac{35}{4}\)
b) 4,5:0,3=2,25:(0,1.x)
\(\frac{45}{10}:\frac{10}{3}=\frac{225}{100}:\left(\frac{1}{10}x\right)\)
\(15=\frac{9}{4}:\left(\frac{1}{10}x\right)\)
\(\frac{9}{4}:\left(\frac{1}{10}x\right)=15\)
\(\frac{1}{10}x=\frac{9}{4}.\frac{1}{15}=\frac{3}{20}\)
\(x=\frac{3}{20}.10=\frac{3}{2}\)
c) 8:\(\left(\frac{1}{4}x\right)=2:0,02\)
8:\(\left(\frac{1}{4}x\right)\)=2:\(\frac{2}{100}\)
8:\(\left(\frac{1}{4}x\right)=2.\frac{100}{2}=100\)
\(\frac{1}{4}x=\frac{8}{100}\)
\(x=\frac{8}{100}.4=\frac{8}{25}\)
d)\(3:2\frac{1}{4}=\frac{3}{4}:\left(6x\right)\)
\(3.\frac{9}{4}=\frac{3}{4}:\left(6x\right)\)
\(\frac{3}{4}:\left(6x\right)=\frac{27}{4}\)
\(6x=\frac{3}{4}.\frac{4}{27}=\frac{1}{9}\)
\(x=\frac{1}{9}.\frac{1}{6}=\frac{1}{54}\)
A) ta có: \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{20}\)
\(\frac{y}{5}=\frac{z}{6}\Rightarrow\frac{y}{20}=\frac{z}{24}\)
=> \(\frac{x}{15}=\frac{y}{20}=\frac{z}{24}=\frac{x+y-z}{15+20-24}=\frac{11}{11}=1\)
=> x=1.15=15
y=1.20=20
z=1.24=24
a] Ta có x/3 = y/4\(\Rightarrow\)x/15 = y/20 ; y/20 = z/24\(\Rightarrow\)x/15 = y/20 = z/24
Theo tính chất của dãy tỉ số bằng nhau ta có
x/15 = y/20 = z/24 = x+y-z/15+20-24 = 11/11 = 1
Từ x/15 = 1\(\Rightarrow\)x/15 = 15/15\(\Rightarrow\)x=15
y/20 = 1\(\Rightarrow\)y/20 = 20/20\(\Rightarrow\)y=20
z/24 = 1\(\Rightarrow\)z/24 = 24/24\(\Rightarrow\)z=24
Vậy x=15;y=20;z=24
Bài 1:\(3^{x+2}-3^x=24\Rightarrow3^x.3^2-3^x=24\Rightarrow3^x.\left(3^2-1\right)=24\Rightarrow3^x.8=24\Rightarrow3^x=3\Rightarrow x=1\)
Bài 2:a,Chọn đáp án C.x0=1
b,Chọn đáp án D\(-\sqrt{2}+\sqrt{5}\) vì \(\sqrt{5}>\sqrt{2}\Rightarrow\left|\sqrt{2}-\sqrt{5}\right|=-\left(\sqrt{2}-\sqrt{5}\right)\)
a, 17- {-x- [-x -(-x) ] } = -16
<=> 17-{-x-[-x+x ] } = -16
<=> 17-{-x-0}=-16
<=> 17-{-x}=-16
<=> 17+x=-16
<=> x=17-(-16)=33
Vậy x = 33
~ học tốt~
nhớ tk nha
\(\Rightarrow17-\left\{-x-\left(-x\right)\right\}=16\)
\(\Rightarrow17-\left\{-x+x\right\}=16\)
\(\Rightarrow17=16\)
Bạn nên viết lại đề bài cho sáng sủa, rõ ràng để người đọc dễ hiểu hơn.
f: =>4(x^2+4x-5)-x^2-7x-10=3(x^2+x-2)
=>4x^2+16x-20-x^2-7x-10-3x^2-3x+6=0
=>6x-24=0
=>x=4
e: =>8x+16-5x^2-10x+4(x^2-x-2)=4-x^2
=>-5x^2-2x+16+4x^2-4x-8=4-x^2
=>-6x+8=4
=>-6x=-4
=>x=2/3
d: =>2x^2+3x^2-3=5x^2+5x
=>5x=-3
=>x=-3/5
b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-12x-5x+20
=>-12x-2=-17x+20
=>5x=22
=>x=22/5
b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-17x+20
=>-12x-2=-17x+20
=>5x=22
=>x=22/5
c: =>24x^2+16x-9x-6-4x^2-16x-7x-28=20x^2-4x+5x-1
=>-16x-34=x-1
=>-17x=33
=>x=-33/17
d: =>2x^2+3x^2-3=5x^2+5x
=>5x=-3
=>x=-3/5
e: =>8x+16-5x^2-10x+4x^2-4x-8=4-x^2
=>-6x+8=4
=>-6x=-4
=>x=2/3
f: =>4(x^2+4x-5)-x^2-7x-10=3x^2+3x-6
=>4x^2+16x-20-4x^2-10x+4=0
=>6x=16
=>x=8/3