Đại số lớp 7

Đề yêu cầu gì em?

Biểu thức trên đạt giá trị lớn nhất khi 2x - 3 > 0 và 2x - 3 nhỏ nhất

2x - 3 > 0

2x > 3

x > 3/2

Mà x là số nguyên

⇒ x ∈ {2; 3; 4; ...}

Mà 2x - 3 nhỏ nhất

⇒ x nhỏ nhất

⇒ x = 2

Vậy x = 2 thì biểu thức đã cho đạt giá trị lớn nhất

|x-0,9|=0

=>x-0,9=0

=>x=0,9

Xét ΔMNP có \(\widehat{M}+\widehat{N}+\widehat{P}=180^0\)

=>\(3\cdot\widehat{P}+2\cdot\widehat{P}+\widehat{P}=180^0\)

=>\(6\cdot\widehat{P}=180^0\)

=>\(\widehat{P}=30^0\)

=>\(\widehat{M}=3\cdot30^0=90^0;\widehat{N}=2\cdot30^0=60^0\)

Bài 2:

Đặt $\frac{x-1}{5}=\frac{y-2}{3}=\frac{z-1}{4}=t$

$\Rightarrow x=5t+1; y=3t+2; z=4t+1$

Khi đó:

$2x-3y-2z=-27$

$\Rightarrow 2(5t+1)-3(3t+2)-2(4t+1)=-27$

$\Rightarrow -7t-6=-27$

$\Rightarrow -7t=-21$

$\Rightarrow t=3$

$x=5t+1=16; y=3t+2=11; z=4.3+1=13$

Bài 1.

 Đặt $6x=10y=15z=t$

$\Rightarrow x=\frac{t}{6}; y=\frac{t}{10}; z=\frac{t}{15}$

$x+y-z=\frac{t}{6}+\frac{t}{10}-\frac{t}{15}=90$

$\Rightarrow t(\frac{1}{6}+\frac{1}{10}-\frac{1}{15})=90$

$\Rightarrow t.\frac{1}{5}=90$

$t=90.5=450$

$x=t:6=75; y=t:10=45; z=t:15=30$

1: 6x=10y=15z

=>\(\dfrac{6x}{60}=\dfrac{10y}{60}=\dfrac{15z}{60}\)

=>\(\dfrac{x}{10}=\dfrac{y}{6}=\dfrac{z}{4}\)

mà x+y-z=90

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{10}=\dfrac{y}{6}=\dfrac{z}{4}=\dfrac{x+y-z}{10+6-4}=\dfrac{90}{12}=7,5\)

=>x=75; y=45; z=30

2: \(\dfrac{x-1}{5}=\dfrac{y-2}{3}=\dfrac{z-1}{4}\)

=>\(\dfrac{2x-2}{10}=\dfrac{3y-6}{9}=\dfrac{2z-2}{8}\)

mà 2x-3y-2z=-27

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{2x-2}{10}=\dfrac{3y-6}{9}=\dfrac{2z-2}{8}=\dfrac{2x-3y-2z-2+6+2}{10-9-8}=\dfrac{-27+6}{-7}=-\dfrac{21}{-7}=3\)

=>\(2x-2=30;3y-6=27;2z-2=24\)

=>2x=32;3y=33;2z=26

=>x=16;y=11;z=13

Bạn tham khảo lời giải tại đây:

https://hoc24.vn/cau-hoi/.8629354073480

|x-2|+|3-x|=1

=>|x-2|+|x-3|=1(1)

TH1: x<2

x<2 nên x-2<0

=>x-2-1<-1<0

=>x-3<0

Phương trình (1) sẽ trở thành:

2-x+3-x=1

=>5-2x=1

=>2x=5-1=4

=>x=2(loại)

TH2: 2<=x<3

x>=2 nên x-2>=0

x<3 nên x-3<0

Phương trình (1) sẽ trở thành:

x-2+3-x=1

=>1=1(luôn đúng)

TH3: x>=3

x>=3 nên x-3>=0

=>x-3+1>=1>0

=>x-2>0

Phương trình (1) sẽ trở thành:

x-2+x-3=1

=>2x-5=1

=>2x=6

=>x=3(nhận)

Vậy: 2<=x<=3

\(\left(x-\dfrac{1}{3}\right)^3=-\dfrac{8}{27}\)

=>\(\left(x-\dfrac{1}{3}\right)^3=\left(-\dfrac{2}{3}\right)^3\)

=>\(x-\dfrac{1}{3}=-\dfrac{2}{3}\)

=>\(x=-\dfrac{2}{3}+\dfrac{1}{3}=-\dfrac{1}{3}\)

Bài 4:

a: \(\sqrt{6^2+8^2}-3\sqrt{25}\)

\(=\sqrt{36+64}-3\cdot5\)

\(=\sqrt{100}-15=10-15=-5\)

b: \(\left(-5\dfrac{1}{2}\right)\left(-\dfrac{1}{2}\right)-\sqrt{\dfrac{4}{3^2}}\cdot\left(-\dfrac{2}{3}\right)\)

\(=\left(-\dfrac{11}{2}\right)\left(-\dfrac{1}{2}\right)-\dfrac{2}{3}\cdot\left(-\dfrac{2}{3}\right)\)

\(=\dfrac{11}{4}+\dfrac{4}{9}=\dfrac{99+16}{36}=\dfrac{115}{36}\)

c: \(\sqrt{16}\cdot\sqrt{4}-\sqrt{25}+2\sqrt{49}\)

\(=4\cdot2-5+2\cdot7\)
\(=8-5+14=22-5=17\)

d: \(\dfrac{1}{\sqrt{36}}+\dfrac{\sqrt{25}}{6}-\sqrt{0,81}\)

\(=\dfrac{1}{6}+\dfrac{5}{6}-0,9\)

=1-0,9

=0,1

e: \(-\dfrac{\sqrt{9}}{16}+\dfrac{5}{\sqrt{36}}\)

\(=-\dfrac{3}{16}+\dfrac{5}{6}\)

\(=\dfrac{-9+40}{48}=\dfrac{31}{48}\)

f: \(\dfrac{\sqrt{9}}{8}\cdot\dfrac{16}{\sqrt{225}}-\dfrac{3}{4\sqrt{4}}\cdot\dfrac{2}{5\sqrt{3^2}}\)

\(=\dfrac{3}{8}\cdot\dfrac{16}{15}-\dfrac{3}{4\cdot2}\cdot\dfrac{2}{5\cdot3}\)

\(=\dfrac{3}{15}\cdot\dfrac{16}{8}-\dfrac{1}{20}\)

\(=\dfrac{1}{5}\cdot2-\dfrac{1}{20}=\dfrac{2}{5}-\dfrac{1}{20}=\dfrac{7}{20}\)

Bài 4:

a) \(\sqrt{6^2+8^2}-3\sqrt{25}\)

\(=\sqrt{36+64}-3\cdot5\)

\(=\sqrt{100}-15\)

\(=10-15\)

\(=-5\)

b) \(\left(-5\dfrac{1}{2}\right)\cdot\left(-\dfrac{1}{2}\right)-\sqrt{\dfrac{4}{3^2}}\cdot\left(-\dfrac{2}{3}\right)\)

\(=\left(-\dfrac{11}{2}\right)\cdot\left(-\dfrac{1}{2}\right)-\dfrac{\sqrt{4}}{\sqrt{3^2}}\cdot\left(-\dfrac{2}{3}\right)\)

\(=\dfrac{11}{4}-\dfrac{2}{3}\cdot\left(-\dfrac{2}{3}\right)\)

\(=\dfrac{11}{4}+\dfrac{4}{9}\)

\(=\dfrac{115}{36}\)

c) \(\sqrt{16}\cdot\sqrt{4}-\sqrt{25}+2\sqrt{49}\)

\(=4\cdot2-5+2\cdot7\)

\(=8-5+14\)

\(=17\)

d) \(\dfrac{1}{\sqrt{36}}+\dfrac{\sqrt{25}}{6}-\sqrt{0,81}\)

\(=\dfrac{1}{6}+\dfrac{5}{6}-0,9\)

\(=1-0,9\)

\(=0,1\)

e) \(-\dfrac{\sqrt{9}}{16}+\dfrac{5}{\sqrt{36}}\)

\(=-\dfrac{3}{16}+\dfrac{5}{6}\)

\(=\dfrac{31}{48}\)

f) \(\dfrac{\sqrt{9}}{8}\cdot\dfrac{16}{\sqrt{225}}-\dfrac{3}{4\sqrt{4}}\cdot\dfrac{2}{5\cdot\sqrt{3^2}}\)

\(=\dfrac{3}{8}\cdot\dfrac{16}{15}-\dfrac{3}{4\cdot2}\cdot\dfrac{2}{5\cdot3}\)

\(=\dfrac{3\cdot16}{8\cdot15}-\dfrac{3\cdot2}{4\cdot2\cdot5\cdot3}\)

\(=\dfrac{2}{5}-\dfrac{1}{20}\)

\(=\dfrac{7}{20}\)

a) \(2,4-3x\cdot0,5=0,9\)

\(\Rightarrow3x\cdot0,5=2,4-0,9\)

\(\Rightarrow3x\cdot0,5=0,5\)

\(\Rightarrow3x=1\)

\(\Rightarrow x=\dfrac{1}{3}\)

b) \(\left|8,8x-50\right|:0,4=51\)

\(\Rightarrow\left|8,8x-50\right|=20,4\)

TH1: \(8,8x-50=20,4\Rightarrow8,8x=70,4\Rightarrow x=8\)

TH2: \(8,8x-50=-20,4\Rightarrow8,8x=29,6\Rightarrow x=\dfrac{37}{11}\)

c) \(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\left|-\dfrac{3}{4}\right|\)

\(\Rightarrow\dfrac{3}{4}\left|x-1\right|=\dfrac{3}{4}\)

\(\Rightarrow\left|x-1\right|=1\)

TH1: \(x-1=1\Rightarrow x=2\)

TH2: \(x-1=-1\Rightarrow x=0\) 

d) \(\left(3x-4\right)^2=\left(-\dfrac{3}{4}\right)^2\)

TH1: \(3x-4=-\dfrac{3}{4}\)

\(\Rightarrow3x=\dfrac{13}{4}\Rightarrow x=\dfrac{13}{12}\)

TH2: \(3x-4=\dfrac{3}{4}\)

\(\Rightarrow3x=\dfrac{19}{4}\Rightarrow x=\dfrac{19}{12}\)

e) \(\left(\dfrac{1}{3}\right)^{2x-1}=3^5\)

\(\Rightarrow\left(\dfrac{1}{3}\right)^{2x-1}=\left[\left(\dfrac{1}{3}\right)^{-1}\right]^5\)

\(\Rightarrow\left(\dfrac{1}{3}\right)^{2x-1}=\left(\dfrac{1}{3}\right)^{-5}\)

\(\Rightarrow2x-1=-5\)

\(\Rightarrow2x=-4\)

\(\Rightarrow x=-2\) 

f) \(\left|5-3x\right|+\dfrac{2}{3}=\dfrac{1}{6}\)

\(\Rightarrow\left|5-3x\right|=-\dfrac{1}{2}\)

TH1: \(5-3x=-\dfrac{1}{2}\Rightarrow3x=\dfrac{11}{2}\Rightarrow x=\dfrac{11}{6}\)

TH2: \(5-3x=\dfrac{1}{2}\Rightarrow3x=\dfrac{9}{2}\Rightarrow x=\dfrac{3}{2}\)