Đại số lớp 7

i don no

hepl me

\(\dfrac{x}{21}=\dfrac{2}{7}-\dfrac{4}{3}\)

\(\dfrac{x}{21}=\dfrac{-22}{21}\)

\(\Leftrightarrow x=-22\)

Vậy .................

Ta có :

\(\left|a+1\right|+\left|b-2\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left|a+1\right|=0\\\left|b-2\right|=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+1=0\\b-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=-1\\b=2\end{matrix}\right.\) \(\left(tm\right)\)

Vậy \(\left\{{}\begin{matrix}a=-1\\b=2\end{matrix}\right.\) là giá trị cần tìm

vì / a+1/ \(\ge\)0 , / b-2/ \(\ge\)0 nên \(\left\{{}\begin{matrix}a+1=0\\b-2=0\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}a=-1\\b=2\end{matrix}\right.\)

1.

a) \(\left|a+1\right|+\left|b-2\right|=0\)

Nhận xét : Do \(\left|a+1\right|\ge0\) với \(\forall a\)

\(\left|b-2\right|\ge0\) với \(\forall b\)

Nên \(\left|a+1\right|+\left|b-2\right|=0\Leftrightarrow\left\{{}\begin{matrix}\left|a+1\right|=0\\\left|b-2\right|=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+1=0\\b-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-1\\b=2\end{matrix}\right.\)

Vậy a = -1, b = 2 t/m đề bài

tik mik nhé !!!

1) Theo đề bài ta có:

\(\dfrac{a}{\dfrac{2}{5}}=\dfrac{b}{\dfrac{3}{4}}=\dfrac{c}{\dfrac{1}{6}}\)

\(\Rightarrow\dfrac{a^2}{\sqrt{\dfrac{2}{5}}}=\dfrac{b^2}{\sqrt{\dfrac{3}{4}}}=\dfrac{c^2}{\sqrt{\dfrac{1}{6}}}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a^2}{\sqrt{\dfrac{2}{5}}}=\dfrac{b^2}{\sqrt{\dfrac{3}{4}}}=\dfrac{c^2}{\sqrt{\dfrac{1}{6}}}\)

\(=\dfrac{a^2+b^2+c^2}{\sqrt{\dfrac{2}{5}}+\sqrt{\dfrac{3}{4}}+\sqrt{\dfrac{1}{6}}}\)

\(=\dfrac{24309}{1,906...}\)

Đến đây thấy đề sai:v

2) Gọi tuổi của 3 anh em lần lượt là \(a;b;c\)

Theo đề bài ta có:

\(\dfrac{3}{4}a=\dfrac{2}{3}b=\dfrac{1}{2}c\)

\(\Rightarrow\left\{{}\begin{matrix}b=\dfrac{3}{4}a:\dfrac{2}{3}=\dfrac{9}{8}a\\c=\dfrac{3}{4}a:\dfrac{1}{2}=\dfrac{3}{4}a\end{matrix}\right.\)

\(\Rightarrow a+\dfrac{9}{8}a+\dfrac{3}{4}a=58\)

\(\Rightarrow\dfrac{22}{8}a=58\)

\(a=\dfrac{232}{11}\)

cả 2 câu là đề sai hay mk tính sai,chẳng hiểu j

Bài 1:

Ta có:

\(a:b:c=\dfrac{2}{5}:\dfrac{3}{4}:\dfrac{1}{6}\)

\(\Rightarrow\dfrac{a}{\dfrac{2}{5}}=\dfrac{b}{\dfrac{3}{4}}=\dfrac{c}{\dfrac{1}{6}}\Rightarrow\dfrac{a^2}{\dfrac{4}{25}}=\dfrac{b^2}{\dfrac{9}{16}}=\dfrac{c^2}{\dfrac{1}{36}}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\dfrac{a^2}{\dfrac{4}{25}}=\dfrac{b^2}{\dfrac{9}{16}}=\dfrac{c^2}{\dfrac{1}{36}}=\dfrac{a^2+b^2+c^2}{\dfrac{4}{25}+\dfrac{9}{16}+\dfrac{1}{36}}\)

\(=\dfrac{24309}{\dfrac{2701}{3600}}=32400\)

\(\Rightarrow\left\{{}\begin{matrix}a^2=5184\\b^2=18225\\c^2=900\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\pm72\\b=\pm135\\c=\pm30\end{matrix}\right.\)

Vậy...........

Chúc bạn học tốt!!!

a) \(\dfrac{5}{2}+x=-3,5\)

\(\Leftrightarrow2,5+x=-3,5\)

\(\Leftrightarrow x=-3,5-2,5\)

\(\Leftrightarrow x=-6\)

Vậy ..................

b) \(-2x+5=23\)

\(-2x=23-5\)

\(-2x=18\)

\(x=18:\left(-2\right)\)

\(x=-9\)

Vậy .................

- Bạn thấy chữ số thứ 3 <5 thì chuyển thành 0

thấy chữ só thứ 3 > 5 thì lấy chữ số thứ 2 cộng thêm 1 rồi chuyển chữ số thứ 3 thành 0

Neeud chữ số thứ 3 =5 thì bn làm theo cách nào cx đc

Bài 11 các bạn nhé. Cảm ơn các bạn.

1 câu thui nhé

(x-1)³ = -8 = (-2)³

x-1 = -2 => x = -2 +1 = -1

( x - 1 ) ³ = -8

=> ( x - 1 ) ³ = ( - 2 )³

=> x - 1 = 2 => x = 3 .

2^{2x - 3} - 2 . 5² = 5² . 3

=> 5² . 3 + 2 . 5² = 5² . 5

=> 5² . 5 = 2 ^{2x - 3}

=> 125 = 2^{2x - 3}

\(\left(x-1\right)^3=-8\)

\(\Rightarrow\left(x-1\right)^3=-2^3\)

\(\Rightarrow x-1=-2\)

\(\Rightarrow x=-1\)

\(2^{2x-3}-2.5^2=5^2.3\)

\(\Rightarrow2^{2x}:2^3-2.25=25.3\)

\(\Rightarrow2^{2x}:8-50=75\)

\(\Rightarrow2^{2x}:8=125\)

\(\Rightarrow2^{2x}=1000\)

Đến đây thì thấy bí

\(M=2x^2-5x+3\)

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

\(=2\left(x^2-\dfrac{5}{4}x-\dfrac{5}{4}x+\dfrac{3}{2}\right)\)

\(=2\left[x\left(x-\dfrac{5}{4}\right)-\dfrac{5}{4}\left(x-\dfrac{5}{4}\right)-\dfrac{1}{16}\right]\)

\(=2\left[\left(x-\dfrac{5}{4}\right)^2-\dfrac{1}{16}\right]\)

\(=2\left(x-\dfrac{5}{4}\right)^2-\dfrac{1}{8}\)

Khi đó: \(2\left(x-\dfrac{5}{4}\right)^2=\dfrac{1}{8}\)

\(\Rightarrow\left(x-\dfrac{5}{4}\right)^2=\left(\dfrac{1}{4}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{5}{4}=\dfrac{1}{4}\\x-\dfrac{5}{4}=-\dfrac{1}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=1\end{matrix}\right.\).

Vậy ...

2x² - 5x + 3 = 0

<=> 2x² - 2x - 3x + 3 = 0

<=> 2x(x - 1) - 3(x - 1) = 0

<=> (x - 1)(2x - 3) = 0

<=> x - 1 = 0 hoặc 2x - 3 = 0

<=> x = 1 hoặc x = 1,5

\(\left|3x-1\right|=x+2\)

\(\Rightarrow\left[{}\begin{matrix}3x-1=x+2\left(đk:x\ge\dfrac{1}{3}\right)\\-3x+1=x+2\left(đk:x< \dfrac{1}{3}\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=x+3\Rightarrow x=\dfrac{2}{3}\left(TM\right)\\-3x=x-1\Rightarrow x=\dfrac{1}{4}\left(TM\right)\end{matrix}\right.\)

\(\left|x-2\right|+\left|3x-1\right|=0\)

\(\left\{{}\begin{matrix}\left|x-2\right|\ge0\\\left|3x-1\right|\ge0\end{matrix}\right.\)

\(\Rightarrow\left|x-2\right|+\left|3x-1\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|x-2\right|=0\Rightarrow x=2\\\left|3x-1\right|=0\Rightarrow3x=1\Rightarrow x=\dfrac{1}{3}\end{matrix}\right.\)

Vì \(2\ne\dfrac{1}{3}\Rightarrow x\in\varnothing\)

\(\left|2x-1\right|=\left|x+1\right|\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=x+1\\-2x-1=-x-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=x+2\Rightarrow x=2\\-2x=-x-2\Rightarrow x=2\end{matrix}\right.\)

a) Ta có :

\(\left|3x-1\right|=x+2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+2=3x-1\\x+2=-\left(3x-1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3+1=3x-x\\x+3x=-2+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=4\\4x=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{-1}{4}\end{matrix}\right.\)

Vậy ......................

b) Ta có :

\(\left|x-2\right|+\left|3x-1\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left|x-2\right|=0\\\left|3x-1\right|=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy ..................

c) Ta có :

\(\left|2x-1\right|=\left|x+1\right|\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=x+1\\2x-1=-x-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-x=1+1\\2x+x=-1+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\3x=0\Leftrightarrow x=0\end{matrix}\right.\)

Vậy ..............

C.

\(\left|2x-1\right|=\left|x+1\right|\\ \Leftrightarrow\left(2x-1\right)^2=\left(x+1\right)^2\\ \Leftrightarrow\left(2x-1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(2x-1+x+1\right)\left(2x-1-x-1\right)=0\\ \Leftrightarrow3x\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

vậy...